NCERT Solutions Class 12 maths Chapter-7 (Integrals)Exercise 7.3

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-7 (Integrals)Exercise 7.3 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.

We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 7.3

Q1. Find the integrals of the functions :

\sin^{2} (2 x + 5)

Answer.sin 2 (2 x + 5) = 1 - cos 2 (2 x + 5) 2 = 1 - cos (4 x + 10) 2 \Rightarrow \int sin 2 (2 x + 5) d x = \int 1 - cos (4 x + 10) 2 d x = 12 \int 1 d x - 12 \int cos (4 x + 10) = 12 x - 12 (sin (4 x + 10) 4) C + = 12 x - 18 sin (4 x + 10) + C

Q2. Find the integrals of the functions : sin 3 x cos 4 x

Answer. It is known that,sin A cos B = 12 {sin (A + B) + sin (A - B)} ∴ \int sin 3 x cos 4 x d x = 12 \int {sin (3 x + 4 x) + sin (3 x - 4 x)} d x = 12 \int {sin 7 x + sin (- x)} d x = 12 \int {sin 7 x - sin x} d x \begin{aligned} \sin A \cos B & = \frac{1}{2} {\sin (A + B) + \sin (A - B)} \\ ∴ \int \sin 3 x \cos 4 x d x & = \frac{1}{2} \int {\sin (3 x + 4 x) + \sin (3 x - 4 x)} d x \\ = \frac{1}{2} \int {\sin 7 x + \sin (- x)} d x \\ = \frac{1}{2} \int {\sin 7 x - \sin x} d x \end{aligned}= 12 \int sin 7 x d x - 12 \int sin x d x = 12 (- cos 7 x 7) - 12 (- cos x) + C = - cos 7 x 14 + cos x 2 + C

Q3. Find the integrals of the functions : cos 2x cos 4x cos 6x

Answer.It is known that, cos A cos B = 12 {cos (A + B) + cos (A - B)} ∴ \int cos 2 x (cos 4 x cos 6 x) d x = \int cos 2 x [12 {cos (4 x + 6 x) + cos (4 x - 6 x)}] d x = 12 \int {cos 2 x cos 10 x + cos 2 x cos (- 2 x)} d x = 12 \int {cos 2 x cos 10 x + cos 2 2 x} d x \begin{aligned}  \end{aligned} It is known that, & \cos A \cos B = \frac{1}{2} {\cos (A + B) + \cos (A - B)} ∴ \int \cos 2 x (\cos 4 x \cos 6 x) d x & = \int \cos 2 x [\frac{1}{2} {\cos (4 x + 6 x) + \cos (4 x - 6 x)}] d x = \frac{1}{2} \int {\cos 2 x \cos 10 x + \cos 2 x \cos (- 2 x)} d x = \frac{1}{2} \int {\cos 2 x \cos 10 x + \cos^{2} 2 x} d x= 12 \int [{12 cos (2 x + 10 x) + cos (2 x - 10 x)} + (1 + cos 4 x 2)] d x = 14 \int (cos 12 x + cos 8 x + 1 + cos 4 x) d x = 14 [sin 12 x 12 + sin 8 x 8 + x + sin 4 x 4] + C

Q4. Find the integrals of the function : $\sin^{3} (2 x + 1)$

Answer.Let I = \int sin 3 (2 x + 1) \Rightarrow \int sin 3 (2 x + 1) d x = \int sin 2 (2 x + 1) \cdot sin (2 x + 1) d x \begin{array}{l}  \end{array} Let I = \int \sin^{3} (2 x + 1) \Rightarrow \int \sin^{3} (2 x + 1) d x = \int \sin^{2} (2 x + 1) \cdot \sin (2 x + 1) d x= \int (1 - cos 2 (2 x + 1)) sin (2 x + 1) d x Let cos (2 x + 1) = t \begin{matrix} = \int (1 - \cos^{2} (2 x + 1)) \sin (2 x + 1) d x \\ Let \cos (2 x + 1) = t \end{matrix}\Rightarrow - 2 sin (2 x + 1) d x = d t \Rightarrow sin (2 x + 1) d x = - d t 2 \begin{array}{l} \Rightarrow - 2 \sin (2 x + 1) d x = d t \\ \Rightarrow \sin (2 x + 1) d x = \frac{- d t}{2} \end{array}\Rightarrow I = - 1 2 \int (1 - t 2) d t = - 1 2 {t - t 3 3} = - 1 2 {cos (2 x + 1) - cos 3 (2 x + 1) 3} = - cos (2 x + 1) 2 + cos 3 (2 x + 1) 6 + C

Q5. Find the integrals of the function : $\sin^{3} x \cos^{3} x$

Answer.Let I = \int sin 3 x cos 3 x \cdot d x = \int cos 3 x \cdot sin 2 x \cdot sin x \cdot d x = \int cos 3 x (1 - cos 2 x) sin x \cdot d x Let cos x = t \Rightarrow - sin x \cdot d x = d t \Rightarrow I = - \int t 3 (1 - t 2) d t \begin{aligned}  \end{aligned} Let I = \int \sin^{3} x \cos^{3} x \cdot d x = \int \cos^{3} x \cdot \sin^{2} x \cdot \sin x \cdot d x = \int \cos^{3} x (1 - \cos^{2} x) \sin x \cdot d x Let \cos x = t \Rightarrow - \sin x \cdot d x = d t \Rightarrow I = - \int t^{3} (1 - t^{2}) d t= - \int (t 3 - t 5) d t = - {t 4 4 - t 6 6} + C = - {cos 4 x 4 - cos 6 x 6} + C = cos 6 x 6 - cos 4 x 4 + C

Q6. Find the integrals of the function : sin x sin 2x sin 3x

Answer. It is known that,sin A sin B = 12 {cos (A - B) - cos (A + B)} ∴ \int sin x sin 2 x sin 3 x d x = \int [sin x \cdot 12 {cos (2 x - 3 x) - cos (2 x + 3 x)}] d x = 12 \int (sin x cos (- x) - sin x cos 5 x) d x \begin{matrix} \sin A \sin B = \frac{1}{2} {\cos (A - B) - \cos (A + B)} \\ ∴ \int \sin x \sin 2 x \sin 3 x d x = \int [\sin x \cdot \frac{1}{2} {\cos (2 x - 3 x) - \cos (2 x + 3 x)}] d x \\ = \frac{1}{2} \int (\sin x \cos (- x) - \sin x \cos 5 x) d x \end{matrix}= 12 \int (sin x cos x - sin x cos 5 x) d x = 12 \int sin 2 x 2 d x - 12 \int sin x cos 5 x d x = 14 [- cos 2 x 2] - 12 \int {12 sin (x + 5 x) + sin (x - 5 x)} d x \begin{array}{l} = \frac{1}{2} \int (\sin x \cos x - \sin x \cos 5 x) d x \\ = \frac{1}{2} \int \frac{\sin 2 x}{2} d x - \frac{1}{2} \int \sin x \cos 5 x d x \\ = \frac{1}{4} [\frac{- \cos 2 x}{2}] - \frac{1}{2} \int {\frac{1}{2} \sin (x + 5 x) + \sin (x - 5 x)} d x \end{array}= - cos 2 x 8 - 14 \int (sin 6 x + sin (- 4 x)) d x = - cos 2 x 8 - 14 [- cos 6 x 3 + cos 4 x 4] + C = - cos 2 x 8 - 18 [- cos 6 x 3 + cos 4 x 2] + C = 18 [cos 6 x 3 - cos 4 x 2 - cos 2 x] + C

Q7. Find the integrals of the function : sin 4x sin 8x

Answer. It is known that,sin A sin B = 12 cos (A - B) - cos (A + B) \sin A \sin B = \frac{1}{2} \cos (A - B) - \cos (A + B)∴ \int sin 4 x sin 8 x d x = \int {12 cos (4 x - 8 x) - cos (4 x + 8 x)} d x = 12 \int (cos (- 4 x) - cos 12 x) d x = 12 \int (cos 4 x - cos 12 x) d x = 12 [sin 4 x 4 - sin 12 x 12]

Q8. Find the integrals of the function : 1 - cos x 1 + cos x

Answer.1 - cos x 1 + cos x = 2 sin 2 2 2 cos 2 x 2 sin 2 x 2 = 1 - cos x and 2 cos 2 x 2 = 1 + cos x] \frac{1 - \cos x}{1 + \cos x} = \frac{2 \sin^{2} 2}{2 \cos^{2} x} 2 \sin^{2} \frac{x}{2} = 1 - \cos x and 2 \cos^{2} \frac{x}{2} = 1 + \cos x]= tan 2 x 2 = (sec 2 x 2 - 1) ∴ \int 1 - cos x 1 + cos x d x = \int (sec 2 x 2 - 1) d x \begin{aligned} = \tan^{2} \frac{x}{2} \\ = (\sec^{2} \frac{x}{2} - 1) \\ ∴ \int \frac{1 - \cos x}{1 + \cos x} d x & = \int (\sec^{2} \frac{x}{2} - 1) d x \end{aligned}= [x - 21 - x] + C = 2 tan x 2 - x + C

Q9. Find the integrals of the function : cos x 1 + cos x

Answer.

\frac{\cos x}{1 + \cos x} = \frac{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}}{2 \cos^{2} \frac{x}{2}} [\cos x = \cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2} and \cos x = 2 \cos^{2} \frac{x}{2} - 1]

\begin{aligned} = & \frac{1}{2} [1 - \tan^{2} \frac{x}{2}] \\ ∴ \frac{\cos x}{1 + \cos x} d x & = \frac{1}{2} \int (1 - \tan^{2} \frac{x}{2}) d x \\ = \frac{1}{2} \int (1 - \sec^{2} \frac{x}{2} + 1) d x \end{aligned}

\begin{array}{l} = \frac{1}{2} [2 x - \frac{\tan \frac{x}{2}}{\frac{1}{2}}] + C \\ = x - \tan \frac{x}{2} + C \end{array}

Q10. Find the integrals of the function : sin 4 x

Answer.sin 4 x = sin 2 x sin 2 x = (1 - cos 2 x 2) (1 - cos 2 x 2) = 14 (1 - cos 2 x) 2 = 14 [1 + cos 2 2 x - 2 cos 2 x] = 14 [1 + (1 + cos 4 x 2) - 2 cos 2 x] \begin{aligned} \sin^{4} x & = \sin^{2} x \sin^{2} x \\ = (\frac{1 - \cos 2 x}{2}) (\frac{1 - \cos 2 x}{2}) \\ = \frac{1}{4} (1 - \cos 2 x)^{2} \\ = \frac{1}{4} [1 + \cos^{2} 2 x - 2 \cos 2 x] \\ = \frac{1}{4} [1 + (\frac{1 + \cos 4 x}{2}) - 2 \cos 2 x] \end{aligned}= 14 [1 + 12 + 12 cos 4 x - 2 cos 2 x] = 14 [32 + 12 cos 4 x - 2 cos 2 x] \begin{aligned} = \frac{1}{4} [1 + \frac{1}{2} + \frac{1}{2} \cos 4 x - 2 \cos 2 x] \\ = \frac{1}{4} [\frac{3}{2} + \frac{1}{2} \cos 4 x - 2 \cos 2 x] \end{aligned}∴ \int sin 4 x d x = 14 \int [32 + 12 cos 4 x - 2 cos 2 x] d x = 14 [32 x + 12 (sin 4 x 4) - 2 sin 2 x 2] + C = 18 [3 x + sin 4 x 4 - 2 sin 2 x] + C = 3 x 8 - 14 sin 2 x + 132 sin 4 x + C

Q11. Find the integrals of the function : cos 4 2 x

Answer.cos 4 2 x = (cos 2 2 x) 2 = (1 + cos 4 x 2) 2 = 14 [1 + cos 2 4 x + 2 cos 4 x] = 14 [1 + (1 + cos 8 x 2) + 2 cos 4 x] \begin{aligned} \cos^{4} 2 x & = {(\cos^{2} 2 x)}^{2} \\ = {(\frac{1 + \cos 4 x}{2})}^{2} \\ = \frac{1}{4} [1 + \cos^{2} 4 x + 2 \cos 4 x] \\ = \frac{1}{4} [1 + (\frac{1 + \cos 8 x}{2}) + 2 \cos 4 x] \end{aligned}= 14 [1 + 12 + cos 8 x 2 + 2 cos 4 x] = 14 [32 + cos 8 x 2 + 2 cos 4 x] \begin{aligned} = \frac{1}{4} [1 + \frac{1}{2} + \frac{\cos 8 x}{2} + 2 \cos 4 x] \\ = \frac{1}{4} [\frac{3}{2} + \frac{\cos 8 x}{2} + 2 \cos 4 x] \end{aligned}∴ \int cos 4 2 x d x = \int (38 + cos 8 x 8 + cos 4 x 2) d x = 38 x + sin 8 x 64 + sin 4 x 8 + C

Q12. Find the integrals of the function :sin 2 x 1 + cos x

Answer.sin 2 x 1 + cos x = (2 sin x 2 cos x 2) 2 2 cos 2 x 2 [sin x = 2 sin x 2 cos x 2; cos x = 2 cos 2 x 2 - 1] \frac{\sin^{2} x}{1 + \cos x} = \frac{{(2 \sin \frac{x}{2} \cos \frac{x}{2})}^{2}}{2 \cos^{2} \frac{x}{2}} [\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}; \cos x = 2 \cos^{2} \frac{x}{2} - 1]= 4 sin 2 x 2 cos 2 x 2 2 cos 2 x 2 = 2 sin 2 x 2 = 1 - cos x \begin{aligned} = \frac{4 \sin^{2} \frac{x}{2} \cos^{2} \frac{x}{2}}{2 \cos^{2} \frac{x}{2}} \\ = 2 \sin^{2} \frac{x}{2} \\ = 1 - \cos x \end{aligned}∴ \int sin - x 1 + cos x d x = \int (1 - cos x) d x = x - sin x + C

Q13. Find the integrals of the function : $\frac{\cos 2 x - \cos 2 α}{\cos x - \cos α}$

Answer.cos 2 x - cos 2 α cos x - cos α = - 2 sin 2 x + 2 α 2 sin 2 x - 2 α 2 - 2 sin x + α 2 sin x - α 2 [cos C - cos D = - 2 sin C + D 2 sin C - D 2] \frac{\cos 2 x - \cos 2 α}{\cos x - \cos α} = \frac{- 2 \sin \frac{2 x + 2 α}{2} \sin \frac{2 x - 2 α}{2}}{- 2 \sin \frac{x + α}{2} \sin \frac{x - α}{2}} [\cos C - \cos D = - 2 \sin \frac{C + D}{2} \sin \frac{C - D}{2}]= sin (x + α) sin (x - α) sin (x + α 2) sin (x - α 2) = [2 sin (x + α 2) cos (x + α 2)] [2 sin (x - α 2) cos (x - α 2)] sin (x + α 2) sin (x - α 2) \begin{aligned} = \frac{\sin (x + α) \sin (x - α)}{\sin (\frac{x + α}{2}) \sin (\frac{x - α}{2})} \\ = \frac{[2 \sin (\frac{x + α}{2}) \cos (\frac{x + α}{2})] [2 \sin (\frac{x - α}{2}) \cos (\frac{x - α}{2})]}{\sin (\frac{x + α}{2}) \sin (\frac{x - α}{2})} \end{aligned}= 4 cos (x + α 2) cos (x - α 2) = 2 [cos (x + α 2 + x - α 2) + cos x + α 2 - x - α 2] = 2 [cos (x) + cos α] = 2 cos x + 2 cos α \begin{aligned} = 4 \cos (\frac{x + α}{2}) \cos (\frac{x - α}{2}) \\ = 2 [\cos (\frac{x + α}{2} + \frac{x - α}{2}) + \cos \frac{x + α}{2} - \frac{x - α}{2}] \\ = 2 [\cos (x) + \cos α] \\ = 2 \cos x + 2 \cos α \end{aligned}∴ \int cos 2 x - cos 2 α cos x - cos α d x = \int 2 cos x + 2 cos α = 2 [sin x + x cos α] + C

Q14. Find the integrals of the function : cos x - sin x 1 + sin 2 x

Answer.

\frac{\cos x - \sin x}{1 + \sin 2 x} = \frac{\cos x - \sin x}{(\sin^{2} x + \cos^{2} x) + 2 \sin x \cos x}

[\sin^{2} x + \cos^{2} x = 1; \sin 2 x = 2 \sin x \cos x]

\begin{matrix} = \frac{\cos x - \sin x}{(\sin x + \cos x)^{2}} \\ Let \sin x + \cos x = t \\ ∴ (\cos x - \sin x) d x = d t \end{matrix}

\begin{aligned} \Rightarrow \int \frac{\cos x - \sin x}{1 + \sin 2 x} d x & = \int \frac{\cos x - \sin x}{(\sin x + \cos x)^{2}} d x \\ = \int \frac{d t}{t^{2}} \\ = \int t^{- 2} d t \\ = - t^{- 1} + C \\ = - t^{- 1} + C \\ = - \frac{1}{\sin x + \cos x} + C \\ = \frac{1}{\sin x + \cos x} + C \end{aligned}

Q15. Find the integrals of the function : tan 3 2 x sec 2 x

Answer.

\begin{aligned} \tan^{3} 2 x \sec 2 x & = \tan^{2} 2 x \tan 2 x \sec 2 x \\ = (\sec^{2} 2 x - 1) \tan 2 x \sec 2 x \\ = \sec^{2} 2 x \cdot \tan 2 x \sec 2 x - \tan 2 x \sec 2 x \\ ∴ \int \tan^{3} 2 x \sec 2 x d x & = \int \sec^{2} 2 x \tan 2 x \sec 2 x d x - \int \tan 2 x \sec 2 x d x \end{aligned}

\begin{array}{l} = \int \sec^{2} 2 x \tan 2 x \sec 2 x d x - \frac{\sec 2 x}{2} + C \\ Let \sec 2 x = t \\ ∴ 2 \sec 2 x \tan 2 x d x = d t \end{array}

\begin{aligned} ∴ \int \tan^{3} 2 x \sec 2 x d x & = \frac{1}{2} \int t^{2} d t - \frac{\sec 2 x}{2} + C \\ = \frac{t^{3}}{6} - \frac{\sec 2 x}{2} + C \\ = \frac{(\sec 2 x)^{3}}{6} - \frac{\sec 2 x}{2} + C \end{aligned}

Q16. Find the integrals of the function :tan 4 x \tan^{4} x

Answer.tan 4 x = tan 2 x \cdot tan 2 x = (sec 2 x - 1) tan 2 x = sec 2 x tan 2 x - tan 2 x = sec 2 x tan 2 x - (sec 2 x - 1) = sec 2 x tan 2 x - sec 2 x + 1 \begin{array}{l} \tan^{4} x \\ = \tan^{2} x \cdot \tan^{2} x \\ = (\sec^{2} x - 1) \tan^{2} x \\ = \sec^{2} x \tan^{2} x - \tan^{2} x \\ = \sec^{2} x \tan^{2} x - (\sec^{2} x - 1) \\ = \sec^{2} x \tan^{2} x - \sec^{2} x + 1 \end{array}∴ \int tan 4 x d x = \int sec 2 x tan 2 x d x - \int sec 2 x d x + \int 1 \cdot d x = \int sec 2 x tan 2 x d x - tan x + x + C . . . (1) Consider \int sec 2 x tan 2 x d x Let tan x = t \Rightarrow sec 2 x d x = d t \Rightarrow \int sec 2 x tan 2 x d x = \int t 2 d t = t 3 3 = tan 3 x 3 \begin{aligned} ∴ \int \tan^{4} x d x = \int \sec^{2} x \tan^{2} x d x - \int \sec^{2} x d x + \int 1 \cdot d x \\ = \int \sec^{2} x \tan^{2} x d x - \tan x + x + C . . . (1) \\ Consider \int \sec^{2} x \tan^{2} x d x \\ Let \tan x = t \Rightarrow \sec^{2} x d x = d t \\ \Rightarrow \int \sec^{2} x \tan^{2} x d x = \int t^{2} d t = \frac{t^{3}}{3} = \frac{\tan^{3} x}{3} \end{aligned}From equation (1), we obtain \int tan 4 x d x = 13 tan 3 x - tan x + x + C

Q17. Find the integrals of the function :sin 3 x + cos 3 x sin 2 x cos 2 x \sin^{3} x + \cos^{3} x \sin^{2} x \cos^{2} x

Answer.sin 3 x + cos 3 x sin 2 x cos 2 x = sin 3 x sin 2 x cos 2 x + cos 3 x sin 2 x cos 2 x = sin x cos 2 x + cos x sin 2 x = tan x sec x + cot x csc x \begin{aligned} \frac{\sin^{3} x + \cos^{3} x}{\sin^{2} x \cos^{2} x} & = \frac{\sin^{3} x}{\sin^{2} x \cos^{2} x} + \frac{\cos^{3} x}{\sin^{2} x \cos^{2} x} \\ = \frac{\sin x}{\cos^{2} x} + \frac{\cos x}{\sin^{2} x} \\ = \tan x \sec x + \cot x \csc x \end{aligned}∴ \int sin 3 x + cos 3 x sin 2 x cos 2 x d x = \int (tan x sec x + cot x csc x) d x = sec x - csc x + C

Q18. Find the integrals of the function : cos 2 x + 2 sin 2 x cos 2 x

Answer.cos 2 x + 2 sin 2 x cos 2 x \frac{\cos 2 x + 2 \sin^{2} x}{\cos^{2} x}= cos 2 x + (1 - cos 2 x) cos 2 x [cos 2 x = 1 - 2 sin 2 x] = \frac{\cos 2 x + (1 - \cos 2 x)}{\cos^{2} x} [\cos 2 x = 1 - 2 \sin^{2} x]= 1 cos 2 x = sec 2 x ∴ \int cos 2 x + 2 sin 2 x cos 2 x d x = \int sec 2 x d x = tan x + C

Q19. Find the integrals of the function : 1 sin x cos 3 x \frac{1}{\sin x \cos^{3} x}

Answer.1 sin x cos 3 x = sin 2 x + cos 2 x sin x cos 3 x = sin x cos 3 x + 1 sin x cos x = tan x sec 2 x + 1 cos 2 x cos 2 x = tan x sec 2 x + sec 2 x tan x \begin{aligned} \frac{1}{\sin x \cos^{3} x} & = \frac{\sin^{2} x + \cos^{2} x}{\sin x \cos^{3} x} \\ = \frac{\sin x}{\cos^{3} x} + \frac{1}{\sin x \cos x} \\ = \tan x \sec^{2} x + \frac{1 \cos^{2} x}{\cos^{2} x} \\ = \tan x \sec^{2} x + \frac{\sec^{2} x}{\tan x} \end{aligned}∴ \int 1 sin x cos 3 x d x = \int tan x sec 2 x d x + \int sec 2 x tan x d x Let tan x = t \Rightarrow sec 2 x d x = d t \Rightarrow \int 1 sin x cos 3 x d x = \int t d t + \int 1 t d t \begin{array}{l} ∴ \int \frac{1}{\sin x \cos^{3} x} d x = \int \tan x \sec^{2} x d x + \int \frac{\sec^{2} x}{\tan x} d x \\ Let \tan x = t \Rightarrow \sec^{2} x d x = d t \\ \Rightarrow \int \frac{1}{\sin x \cos^{3} x} d x = \int t d t + \int_{t}^{1} d t \end{array}= t 2 2 + | log t | + C = 12 tan 2 x + | log | tan x | + C

Q20. Find the integrals of the function : cos 2 x (cos x + sin x) 2

Answer.cos 2 x (cos x + sin x) 2 = cos 2 x cos 2 x + sin 2 x + 2 sin x cos x = cos 2 x 1 + sin 2 x ∴ \int cos 2 x (cos x + sin x) 2 d x = \int cos 2 x (1 + sin 2 x) d x \begin{array}{l} \frac{\cos 2 x}{(\cos x + \sin x)^{2}} = \frac{\cos 2 x}{\cos^{2} x + \sin^{2} x + 2 \sin x \cos x} = \frac{\cos 2 x}{1 + \sin 2 x} \\ ∴ \int \frac{\cos 2 x}{(\cos x + \sin x)^{2}} d x = \int \frac{\cos 2 x}{(1 + \sin 2 x)} d x \end{array}∴ \int cos 2 x (cos x + sin x) 2 d x = \int cos 2 x (1 + sin 2 x) d x Let 1 + sin 2 x = t \Rightarrow 2 cos 2 x d x = d t ∴ \int cos 2 x (cos x + sin x) 2 d x = 12 \int 1 t d t \begin{array}{l} ∴ \int \frac{\cos 2 x}{(\cos x + \sin x)^{2}} d x = \int \frac{\cos 2 x}{(1 + \sin 2 x)} d x \\ Let 1 + \sin 2 x = t \\ \Rightarrow 2 \cos 2 x d x = d t \\ ∴ \int \frac{\cos 2 x}{(\cos x + \sin x)^{2}} d x = \frac{1}{2} \int_{t}^{1} d t \end{array}= 12 log | t | + C = 12 log | 1 + sin 2 x | + C = 12 log ∣ ∣ (sin x + cos x) 2 ∣ ∣ + C = log | sin x + cos x | + C

Q21. Find the integrals of the function :sin - 1 (cos x) \sin^{- 1} (\cos x)

Answer.sin - 1 (cos x) Let cos x = t Then, sin x = \sqrt 1 - t 2 \begin{array}{l} \sin^{- 1} (\cos x) \\ Let \cos x = t \\ Then, \sin x = \sqrt{1 - t^{2}} \end{array}\Rightarrow (- sin x) d x = d t d x = - d t sin x d x = - d t \sqrt 1 - t 2 ∴ \int sin - 1 (cos x) d x = \int sin - 1 t (- d t \sqrt 1 - t 2) \begin{array}{l} \Rightarrow (- \sin x) d x = d t \\ d x = \frac{- d t}{\sin x} \\ d x = \frac{- d t}{\sqrt{1 - t^{2}}} \\ ∴ \int \sin^{- 1} (\cos x) d x = \int \sin^{- 1} t (\frac{- d t}{\sqrt{1 - t^{2}}}) \end{array}= - \int sin - 1 t \sqrt 1 - t 2 d t Let sin - 1 t = u \Rightarrow 1 \sqrt 1 - t 2 d t = d u ∴ \int sin - 1 (cos x) d x = \int 4 d u \begin{array}{l} = - \int \frac{\sin^{- 1} t}{\sqrt{1 - t^{2}}} d t \\ Let \sin^{- 1} t = u \\ \Rightarrow \frac{1}{\sqrt{1 - t^{2}}} d t = d u \\ ∴ \int \sin^{- 1} (\cos x) d x = \int 4 d u \end{array}= - u 2 2 + C = - (sin 1 t) 2 2 + C = - [sin - 1 (cos x)] 2 2 + C \begin{array}{l} = - \frac{u^{2}}{2} + C \\ = \frac{- {(\sin^{1} t)}^{2}}{2} + C \\ = \frac{- {[\sin^{- 1} (\cos x)]}^{2}}{2} + C \end{array}It is known that, sin - 1 x + cos - 1 x = π 2 ∴ sin - 1 (cos x) = π 2 - cos - 1 (cos x) = (π 2 - x) Substituting in equation (1), we obtain \begin{array}{l} It is known that, \\ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2} \\ ∴ \sin^{- 1} (\cos x) = \frac{π}{2} - \cos^{- 1} (\cos x) = (\frac{π}{2} - x) \\ Substituting in equation (1), we obtain \end{array}\int sin - 1 (cos x) d x = - [π 2 - x] 2 2 + C = - 12 (π 2 2 + x 2 - π x) + C = π x 8 - x 2 2 + 12 π x + C = π x 2 - x 2 2 + (C - π 2 8) = π x 2 - x 2 2 + C 1

Q22. Find the integrals of the function : 1 cos (x - a) cos (x - b)

Answer.1 cos (x - a) cos (x - b) = 1 sin (a - b) [sin (a - b) cos (x - a) cos (x - b)] = 1 sin (a - b) [sin [(x - b) - (x - a)] cos (x - a) cos (x - b)]] = 1 sin (a - b) [sin (x - b) cos (x - a) - cos (x - b) sin (x - a)] sin (a - b) \begin{aligned} \frac{1}{\cos (x - a) \cos (x - b)} & = \frac{1}{\sin (a - b)} [\frac{\sin (a - b)}{\cos (x - a) \cos (x - b)}] \\ = \frac{1}{\sin (a - b)} [\frac{\sin [(x - b) - (x - a)]}{\cos (x - a) \cos (x - b)]}] \\ = \frac{1}{\sin (a - b)} \frac{[\sin (x - b) \cos (x - a) - \cos (x - b) \sin (x - a)]}{\sin (a - b)} \end{aligned}= 1 sin (a - b) [tan (x - b) - tan (x - a)] \Rightarrow \int 1 cos (x - a) cos (x - b) d x = 1 sin (a - b) \int [tan (x - b) - tan (x - a)] d x = 1 sin (a - b) [- log | cos (x - b) | + log cos (x - a) |] = 1 sin (a - b) [log ∣ ∣ ∣ cos (x - a) cos (x - b) ∣ ∣ ∣] + C

Q23. Choose the correct answer \int sin 2 x - cos 2 x sin 2 x cos 2 x d x \int \frac{\sin^{2} x - \cos^{2} x}{\sin^{2} x \cos^{2} x} d x is equal to (A) tan x + cot x + C (B) tan x + cosec x + C (C) - tan x + cot x + C (D) tan x + sec x + C

Answer.\int sin 2 x - cos 2 x sin 2 x cos 2 x d x = \int sin 2 x sin 2 x cos 2 x - cos 2 x sin 2 x cos 2 x) d x = \int (sec 2 x - csc 2 x) d x = tan x + cot x + C Hence, the correct Answer is A

Q24. $\int \frac{e^{x} (1 + x)}{\cos^{2} (e^{x} x)}$ equals $\begin{array}{ll} (A) - \cot (e x^{x}) + C & (B) \tan (x e^{x}) + C \\ (C) & \tan (e^{x}) + C & (D) \cot (e^{x}) + C \end{array}$

Answer.

\int \frac{e^{x} (1 + x)}{\cos^{2} (e^{x} x)}

Let exx = t

\begin{array}{l} \Rightarrow (e^{x} \cdot x + e^{x} \cdot 1) d x = d t \\ e^{x} (x + 1) d x = d t \\ ∴ \int \frac{e^{x} (1 + x)}{\cos^{2} (e^{x} x)} d x = \int \frac{d t}{\cos^{2} t} \end{array}

\begin{aligned} = \int \sec^{2} t d t \\ = \tan t + C \\ = \tan (e^{x} \cdot x) + C \end{aligned}

Hence, the correct Answer is B.

NCERT Solutions Class 12 maths Chapter-7 (Integrals) Exercise 7.3

NCERT Solutions Class 12 maths Chapter-7 (Integrals)Exercise 7.3

Exercise 7.3

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NCERT Solutions Class 12 maths Chapter-7 (Integrals) Exercise 7.3

NCERT Solutions Class 12 maths Chapter-7 (Integrals)Exercise 7.3

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