NCERT Solutions Class 12 Maths Chapter-7 (Integrals)Exercise 7.9

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-7 (Integrals)Exercise 7.9 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 7.9

Q1. Evaluate the definite integral : ${\int }_{-1}^1(x+1)dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_{-1}^4(x+1)dx\\ \int (x+1)dx=\frac{x^2}{2}+x=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $I=F\left(1\right)-F(-1)$ $\begin{array}{c}=(\frac{1}{2}+1)-(\frac{1}{2}-1)\\ =\frac{1}{2}+1-\frac{1}{2}+1\\ =2\end{array}$

Q2. Evaluate the definite integral : ${\int }_x^1\frac{1}{x}dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_2^3\frac{1}{x}dx\\ \int \frac{1}{x}dx=\log |x|=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(3\right)-F\left(2\right)\\ & =\log |3|-\log |2|=\log \frac{3}{2}\end{array}$

Q3. Evaluate the definite integral : ${\int }^2(4x^3-5x^2+6x+9)dx$

Answer. $\begin{array}{cc}\text{Let}I=\int (4x^3-5x^2+6x+9)dx& \\ \int (4x^3-5x^2+6x+9)dx& =4\left(\frac{x^4}{4}\right)-5\left(\frac{x^3}{3}\right)+6\left(\frac{x^2}{2}\right)+9\left(x\right)\\ & =x^4-\frac{5x^3}{3}+3x^2+9x=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(2\right)-F\left(1\right)\\ I& =\{2^4-\frac{5\cdot (2{)}^3}{3}+3(2{)}^2+9\left(2\right)\}-\left\{\right(1{)}^4-\frac{5(1{)}^3}{3}+3(1{)}^2+9(1\left)\right\}\\ & =(16-\frac{40}{3}+12+18)-(1-\frac{5}{3}+3+9)\\ & =16-\frac{40}{3}+12+18-1+\frac{5}{3}-3-9\\ & =33-\frac{35}{3}\\ & =\frac{99-35}{3}\\ & =\frac{64}{3}\end{array}$

Q4. Evaluate the definite integral : ${\int }_0^x\sin 2xdx$

Answer. $\begin{array}{c}\text{Let}I={\int }_0^{\pi }\sin 2xdx\\ \int \sin 2xdx=\left(\frac{-\cos 2x}{2}\right)=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(\frac{\pi }{4}\right)-F\left(0\right)\\ & =-\frac{1}{2}[\cos 2\left(\frac{1}{4}\right)-\cos 0]\\ & =-\frac{1}{2}[\cos \left(\frac{-}{2}\right)-\cos 0]\\ & =-\frac{1}{2}[0-1]\\ & =\frac{1}{2}\end{array}$

Q5. Evaluate the definite integral : ${\int }^{\frac{x}{2}}\cos 2xdx$

Answer. $\begin{array}{c}\text{Let}I={\int }_0^{\frac{\pi }{2}}\cos 2xdx\\ \int \cos 2xdx=\left(\frac{\sin 2x}{2}\right)=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(\frac{\pi }{2}\right)-F\left(0\right)\\ & =\frac{1}{2}[\sin 2\left(\frac{\pi }{2}\right)-\sin 0]\\ & =\frac{1}{2}[\sin \pi -\sin 0]\\ & =\frac{1}{2}[0-0]=0\end{array}$

Q6. Evaluate the definite integral : ${\int }_1^6{e}^{x}dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_4^6{e}^{x}dx\\ \int e^{x}dx=e^x=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(5\right)-F\left(4\right)\\ & =e^5-e^4\\ & =e^4(e-1)\end{array}$

Q7. Evaluate the definite integral : ${\int }_0^x\tan xdx$

Answer. $\begin{array}{c}\text{Let}I={\int }_5^{\pi }\tan xdx\\ \int \tan xdx=-\log |\cos x|=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(\frac{\pi }{4}\right)-F\left(0\right)\\ & =-\log \left|\cos \frac{\pi }{4}\right|+\log |\cos 0|\\ & =-\log |\frac{1}{\sqrt{2}|}+\log |l|\\ & =-\log (2{)}^{-\frac{1}{2}}\\ & =\frac{1}{2}\log 2\end{array}$

Q8. Evaluate the definite integral : ${\int }_6^{\frac{\pi }{4}}\cos \mathrm{ec}xdx$

Answer. $I={\int }_{\pi }^{\frac{\pi }{4}}\cos \mathrm{ec}xdx$ $\int \csc xdx=\log |\csc x-\cot x|=F\left(x\right)$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(\frac{\pi }{4}\right)-F\left(\frac{\pi }{6}\right)\\ & =\log |\csc \frac{\pi }{4}-\cot \frac{\pi }{4}|-\log |\csc \frac{\pi }{6}-\cot \frac{\pi }{6}|\\ & =\log |\sqrt{2}-1|-\log |2-\sqrt{3}|\\ & =\log \left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\end{array}$

Q9. Evaluate the definite integral : ${\int }_0^4\frac{dx}{\sqrt{1-x^2}}$

Answer. $\begin{array}{c}\text{Let}I={\int }_0^4\frac{dx}{\sqrt{1-x^2}}\\ \int \frac{dx}{\sqrt{1-x^2}}={\sin }^{-1}x=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(1\right)-F\left(0\right)\\ & ={\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(0\right)\\ & =\frac{\pi }{2}-0\\ & =\frac{\pi }{2}\end{array}$

Q10. Evaluate the definite integral : $\int \frac{dx}{1+x^2}$

Answer. $\begin{array}{c}\text{Let}I=\int \frac{dx}{1+x^2}\\ \int \frac{dx}{1+x^2}={\tan }^{-1}x=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(1\right)-F\left(0\right)\\ & ={\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(0\right)\\ & =\frac{\pi }{4}\end{array}$

Q11. Evaluate the definite integral : ${\int }_2^3\frac{dx}{x^2-1}$

Answer. $\begin{array}{c}\text{Let}I={\int }_2^3\frac{dx}{x^2-1}\\ \int \frac{dx}{x^2-1}=\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(3\right)-F\left(2\right)\\ & =\frac{1}{2}[\log \left|\frac{3-1}{3+1}\right|-\log |\frac{2-1}{2+1}]\\ & =\frac{1}{2}[\log \left|\frac{2}{4}\right|-\log \frac{1}{3}]\\ & =\frac{1}{2}[\log \frac{1}{2}-\log \frac{1}{3}]\\ & =\frac{1}{2}\left[\log \frac{3}{2}\right]\end{array}$

Q12. Evaluate the definite integral : ${\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx$

Answer. $\begin{array}{c}\text{Let}I={\int }_1^{\frac{\pi }{2}}{\cos }^{2}xdx\\ \int {\cos }^{2}xdx=\int \left(\frac{1+\cos 2x}{2}\right)dx=\frac{x}{2}+\frac{\sin 2x}{4}=\frac{1}{2}(x+\frac{\sin 2x}{2})=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =[F\left(\frac{\pi }{2}\right)-F(0\left)\right]\\ & =\frac{1}{2}[(\frac{\pi }{2}-\frac{\sin \pi }{2})-(0+\frac{\sin 0}{2})]\\ & =\frac{1}{2}[\frac{\pi }{2}+0-0-0]\\ & =\frac{\pi }{4}\end{array}$

Q13. Evaluate the definite integral : ${\int }_2^3\frac{xdx}{x^2+1}$

Answer. $\begin{array}{c}\text{Let}I={\int }_2^3\frac{x}{x^2+1}dx\\ \int \frac{x}{x^2+1}dx=\frac{1}{2}\int \frac{2x}{x^2+1}dx=\frac{1}{2}\log (1+x^2)=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(3\right)-F\left(2\right)\\ & =\frac{1}{2}[\log (1+(3{)}^2)-\log (1+(2{)}^2)]\\ & =\frac{1}{2}\left[\log \right(10)-\log (5\left)\right]\\ & =\frac{1}{2}\log \left(\frac{10}{5}\right)=\frac{1}{2}\log 2\end{array}$

Q14. Evaluate the definite integral : $\int \frac{2x+3}{5x^2+1}dx$

Answer. $I={\int }_0^4\frac{2x+3}{5x^2+1}dx$ $\begin{array}{cc}\int \frac{2x+3}{5x^2+1}dx& =\frac{1}{5}\int \frac{5(2x+3)}{5x^2+1}dx\\ & =\frac{1}{5}\int \frac{10x+15}{5x^2+1}dx\\ & =\frac{1}{5}\int \frac{10x}{5x^2+1}dx+3\int \frac{1}{5x^2+1}dx\end{array}$ $\begin{array}{cc}& =\frac{1}{5}\int \frac{10x}{5x^2+1}dx+3\int \frac{1}{5(x^2+\frac{1}{5})}dx\\ & =\frac{1}{5}\log (5x^2+1)+\frac{3}{5}\cdot \frac{1}{\sqrt{5}}{\tan }^{-1}\frac{x}{\sqrt{5}}\\ & =\frac{1}{5}\log (5x^2+1)+\frac{3}{\sqrt{5}}{\tan }^{-1}\left(\sqrt{5}x\right)\\ & =F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(1\right)-F\left(0\right)\\ & =\left\{\frac{1}{5}\log \right(5+1)+\frac{3}{\sqrt{5}}{\tan }^{-1}(\sqrt{5}\left)\right\}-\left\{\frac{1}{5}\log \right(1)+\frac{3}{\sqrt{5}}{\tan }^{-1}(0\left)\right\}\\ & =\frac{1}{5}\log 6+\frac{3}{\sqrt{5}}{\tan }^{-1}\sqrt{5}\end{array}$

Q15. Evaluate the definite integral : ${\int }_0^{1}x{e}^{x^2}dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_{0}x{e}^{x^2}dx\\ \text{Put}{x}^2=t\Rightarrow 2xdx=dt\\ \text{As}x\to 0,t\to 0\text{and as}x\to 1,t\to 1\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(1\right)-F\left(0\right)\\ & =\frac{1}{2}e-\frac{1}{2}{e}^0\\ & =\frac{1}{2}(e-1)\end{array}$

Q16. Evaluate the definite integral : $\int \frac{5x^2}{x^2+4x+3}$

Answer. Let $I={\int }_i^2\frac{5x^2}{x^2+4x+3}dx$ Dividing $5x^2\text{by}{x}^2+4x+3$ we obtain  $\begin{array}{cc}I& ={\int }_1^2\{5-\frac{20x+15}{x^2+4x+3}\}dx\\ & ={\int }_1^2\{5dx-{\int }_1^2\frac{20x+15}{x^2+4x+3}dx\\ & =[5x{]}_1^2-{\int }_1^2\frac{20x+15}{x^2+4x+3}dx\\ I& =5-I_1,\text{where}I={\int }_1^2\frac{20x+15}{x^2+4x+3}dx\end{array}$ Consider $I_1={\int }_i^2\frac{20x+15}{x^2+4x+8}dx$ $\begin{array}{cc}\text{Consider}{I}_1=& {\int }_1^2\frac{20x+15}{x^2+4x+8}dx\\ \text{Let}20x+15& =A\frac{d}{dx}(x^2+4x+3)+B\\ & =2Ax+(4A+B)\end{array}$ Equating the coefficients of x and constant term, we obtain $\begin{array}{c}A=10\text{and}B=-25\\ \Rightarrow I_1=10\int \frac{2x+4}{x^2+4x+3}dx-25{\int }_1^2\frac{dx}{x^2+4x+3}\\ \text{Let}{x}^2+4x+3=t\\ \Rightarrow (2x+4)dx=dt\end{array}$ $\begin{array}{cc}\Rightarrow I_1& =10\int \frac{dt}{t}-25\int \frac{dx}{(x+2{)}^2-1^2}\\ & =10\log t-25\left[\frac{1}{2}\log \left(\frac{x+2-1}{x+2+1}\right)\right]\\ & ={\left[10\log (x^2+4x+3)\right]}_1^2-25{\left[\frac{1}{2}\log \left(\frac{x+1}{x+3}\right)\right]}_1^2\\ & =[10\log 15-10\log 8]-25[\frac{1}{2}\log \frac{3}{5}-\frac{1}{2}\log \frac{2}{4}]\\ & =\left[10\log \right(5\times 3)-10\log (4\times 2\left)\right]-\frac{25}{2}[\log 3-\log 5-\log 2+\log 4]\end{array}$ $\begin{array}{cc}& =[10\log 5+10\log 3-10\log 4-10\log 2]-\frac{25}{2}[\log 3-\log 5-\log 2+\log 4]\\ & =[10+\frac{25}{2}]\log 5+[-10-\frac{25}{2}]\log 4+[10-\frac{25}{2}]\log 3+[-10+\frac{25}{2}]\log 2\\ & =\frac{45}{2}\log 5-\frac{45}{2}\log 4-\frac{5}{2}\log 3+\frac{5}{2}\log 2\\ & =\frac{45}{2}\log \frac{5}{4}-\frac{5}{2}\log \frac{3}{2}\end{array}$ Substituting the value of I1 in (1), we obtain $\begin{array}{cc}I& =5-[\frac{45}{2}\log \frac{5}{4}-\frac{5}{2}\log \frac{3}{2}]\\ & =5-\frac{5}{2}[9\log \frac{5}{4}-\log \frac{3}{2}]\end{array}$

Q17. Evaluate the definite integral : ${\int }_0^{\pi }(2{\sec }^{2}x+x^3+2)dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_5^{\frac{\pi }{4}}(2{\sec }^{2}x+x^3+2)dx\\ \int (2{\sec }^{2}x+x^3+2)dx=2\tan x+\frac{x^4}{4}+2x=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(\frac{\pi }{4}\right)-F\left(0\right)\\ & =\{(2\tan \frac{\pi }{4}+\frac{1}{4}{\left(\frac{\pi }{4}\right)}^4+2\left(\frac{\pi }{4}\right))-(2\tan 0+0+0\left)\right\}\\ & =2\tan \frac{\pi }{4}+\frac{{\pi }^4}{4^5}+\frac{\pi }{2}\\ & =2+\frac{\pi }{2}+\frac{{\pi }^4}{1024}\end{array}$

Q18. Evaluate the definite integral : ${\int }_0^a({\sin }^2\frac{x}{2}-{\cos }^2\frac{x}{2})dx$

Answer. $\begin{array}{cc}\text{Let}I& ={\int }_0^{\pi }({\sin }^2\frac{x}{2}-{\cos }^2\frac{x}{2})dx\\ & =-{\int }_0^{\pi }({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2})dx\\ & =-{\int }_0^{\pi }\cos xdx\\ \int \cos xdx& =\sin x=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(\pi \right)-F\left(0\right)\\ & =\sin \pi -\sin 0\\ & =0\end{array}$

Q19. Evaluate the definite integral : ${\int }_0^2\frac{6x+3}{x^2+4}dx$

Answer. $\begin{array}{cc}\text{Let}I=& {\int }_0^2\frac{6x+3}{x^2+4}dx\\ \int \frac{6x+3}{x^2+4}dx& =3\int \frac{2x+1}{x^2+4}dx\\ & =3\int \frac{2x}{x^2+4}dx+3\int \frac{1}{x^2+4}dx\\ & =3\log (x^2+4)+\frac{3}{2}{\tan }^{-1}\frac{x}{2}=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(2\right)-F\left(0\right)\\ & =\{3\log (2^2+4)+\frac{3}{2}{\tan }^{-1}\left(\frac{2}{2}\right)\}-\left\{3\log \right(0+4)+\frac{3}{2}{\tan }^{-1}\left(\frac{0}{2}\right)\}\\ & =3\log 8+\frac{3}{2}{\tan }^{-1}1-3\log 4-\frac{3}{2}{\tan }^{-1}0\\ & =3\log 8+\frac{3}{2}\left(\frac{\pi }{4}\right)-3\log 4-0\\ & =3\log \left(\frac{8}{4}\right)+\frac{3\pi }{8}\\ & =3\log 2+\frac{3\pi }{8}\end{array}$

Q20. Evaluate the definite integral : ${\int }_0^1(x{e}^x+\sin \frac{\pi x}{4})dx$

Answer. $I={\int }_0^1(x{e}^x+\sin \frac{\pi x}{4})dx$ $\int (x{e}^x+\sin \frac{\pi x}{4})dx=x\int e^{x}dx-\int \{\left(\frac{d}{dx}x\right)\int e^{x}dx\}dx+\left\{\begin{array}{c}-\cos \frac{\pi x}{4}\\ \frac{\pi }{4}\end{array}\right\}$ $\begin{array}{cc}& =x{e}^x-\int e^{x}dx-\frac{4\pi }{\pi }\cos \frac{x}{4}\\ & =x{e}^x-e^x-\frac{4\pi }{\pi }\cos \frac{x}{4}\\ & =F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(1\right)-F\left(0\right)\\ & =(1.e^1-e^1-\frac{4}{\pi }\cos \frac{\pi }{4})-(0.e^0-e^0-\frac{4}{\pi }\cos 0)\\ & =e-e-\frac{4}{\pi }\left(\frac{1}{\sqrt{2}}\right)+1+\frac{4}{\pi }\\ & =1+\frac{4}{\pi }-\frac{2\sqrt{2}}{\pi }\end{array}$

Q21. Choose the correct answer : ${\int }^{\sqrt{3}}\frac{dx}{1+x^2}$ equals (A) $\frac{\pi }{3}$ (B) $\frac{2\pi }{3}$ (C) $\frac{\pi }{6}$ (D) $\frac{\pi }{12}$

Answer. $\int \frac{dx}{1+x^2}={\tan }^{-1}x=F\left(x\right)$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}{\int }^{\sqrt{3}}\frac{dx}{1+x^2}& =F\left(\sqrt{3}\right)-F\left(1\right)\\ & ={\tan }^{-1}\sqrt{3}-{\tan }^{-1}1\\ & =\frac{\pi }{3}-\frac{\pi }{4}\\ & =\frac{\pi }{12}\end{array}$ Hence, the correct Answer is D.

${\int }_0^{1}x{e}^{x^2}dx$

Q22. Choose the correct answer : ${\int }_0^{\frac{2}{3}}\frac{dx}{4+9x^2}$ equals (A) $\frac{\pi }{6}$ (B) $\frac{\pi }{12}$ (C) $\frac{\pi }{24}$ (D) $\frac{\pi }{4}$

Answer. $\begin{array}{cc}\int \frac{dx}{4+9x^2}& =\int \frac{dx}{(2{)}^2+(3x{)}^2}\\ \text{Put}3x=t\Rightarrow 3dx& =dt\\ \therefore \int \frac{dx}{(2{)}^2+(3x{)}^2}& =\frac{1}{3}\int \frac{dt}{(2{)}^2+t^2}\\ & =\frac{1}{3}\left[\frac{1}{2}{\tan }^{-1}\frac{t}{2}\right]\\ & =\frac{1}{6}{\tan }^{-1}\left(\frac{3x}{2}\right)\\ & =F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}{\int }_0^{\frac{2}{3}}\frac{dx}{4+9x^2}& =F\left(\frac{2}{3}\right)-F\left(0\right)\\ & =\frac{1}{6}{\tan }^{-1}(\frac{3}{2}\cdot \frac{2}{3})-\frac{1}{6}{\tan }^{-1}0\\ & =\frac{1}{6}{\tan }^{-1}1-0\\ & =\frac{1}{6}\times \frac{\pi }{4}\\ & =\frac{\pi }{24}\end{array}$ Hence the correct answer is C

Chapter-7 (integrals)

• Exercise 7.1
• Exercise 7.2
• Exercise 7.3
• Exercise 7.4
• Exercise 7.5
• Exercise 7.6
• Exercise 7.7
• Exercise 7.8
• Exercise 7.10
• Exercise 7.11