NCERT Solutions Class 12 Maths Chapter-7 (Integrals)Exercise 7.11

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-7 (Integrals)Exercise 7.11 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 7.11 

Q1. By using the properties of definite integrals, evaluate the integral. ${\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx$

Answer. $I={\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx$ ……..(1) $\Rightarrow I={\int }_0^{\frac{\pi }{2}}{\cos }^2(\frac{\pi }{2}-x)dx\left({\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right)$ $\Rightarrow I={\int }_0^{\frac{\pi }{2}}{\sin }^{2}xdx$ ………..(2) $\begin{array}{c}\text{Adding}\left(1\right)\text{and}\left(2\right),\text{we obtain}\\ 2I={\int }_b^{\frac{\pi }{2}}({\sin }^{2}x+{\cos }^{2}x)dx\\ \Rightarrow 2I={\int }_0^{\frac{\pi }{2}}1dx\\ \Rightarrow 2I=[x{]}_0^{\frac{\pi }{2}}\\ \Rightarrow 2I=\frac{\pi }{2}\\ \Rightarrow I=\frac{\pi }{4}\end{array}$

Q2. By using the properties of definite integrals, evaluate the integral. ${\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$

Answer. ${\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ Let $I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx.......\left(1\right)$ $\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sin (\frac{\pi }{2}-x)}}{\sqrt{\sin (\frac{\pi }{2}-x)}+\sqrt{\cos (\frac{\pi }{2}-x)}}dx\left({\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right)$ $\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\cos }}{\sqrt{\cos }+\sqrt{\sin x}}dx......\left(2\right)$ $\begin{array}{c}\text{Adding}\left(1\right)\text{and}\left(2\right),\text{we obtain}\\ 2I={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx\\ \Rightarrow 2I={\int }_0^{\frac{\pi }{2}}1dx\\ \Rightarrow 2I=[x{]}_0^{\frac{\pi }{2}}\\ \Rightarrow 2I=\frac{\pi }{2}\\ \Rightarrow I=\frac{\pi }{4}\end{array}$

Q3. By using the properties of definite integrals, evaluate the integral. ${\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{\frac{3}{2}}xdx}{{\sin }^{\frac{3}{2}}x+{\cos }^{\frac{3}{2}}x}$

Q4. By using the properties of definite integrals, evaluate the integral. ${\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{5}xdx}{{\sin }^{5}x+{\cos }^{5}x}$

Answer. $LetI={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{5}x}{{\sin }^{5}x+{\cos }^{5}x}dx......\left(1\right)$ $\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^5(\frac{\pi }{2}-x)}{{\sin }^5(\frac{\pi }{2}-x)+{\cos }^5(\frac{\pi }{2}-x)}dx\left({\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right)$ $\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{5}x}{{\sin }^{5}x+{\cos }^{5}x}dx.....\left(2\right)$ $\begin{array}{c}\text{Adding}\left(1\right)\text{and}\left(2\right),\text{we obtain}\\ 2I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{5}x+{\cos }^{5}x}{{\sin }^{5}x+{\cos }^{5}x}dx\\ \Rightarrow 2I={\int }_0^{\frac{x}{2}}1dx\\ \Rightarrow 2I=[x{]}_0^{\frac{\pi }{2}}\\ \Rightarrow 2I=\frac{\pi }{2}\\ \Rightarrow I=\frac{\pi }{4}\end{array}$

Q5. By using the properties of definite integrals, evaluate the integral. ${\int }_{-5}^5|x+2|dx$

Answer. $I={\int }_{-5}^5|x+2|dx$ It can be seen that (x+2) $\le 0\text{on}[-5,-2]\text{and}(x+2)\ge 0\text{on}[-2,5]$ $\therefore I={\int }_{-5}^{-2}-(x+2)dx+{\int }_{-2}^5(x+2)dx\left({\int }_a^{b}f\right(x)={\int }_a^{c}f(x)+{\int }_c^{b}f(x\left)\right)$ $I=-{[\frac{x^2}{2}+2x]}_{-5}^{-2}+{[\frac{x^2}{2}+2x]}_{-2}^5$ $=-[\frac{(-2{)}^2}{2}+2(-2)-\frac{(-5{)}^2}{2}-2(-5\left)\right]+[\frac{(5{)}^2}{2}+2(5)-\frac{(-2{)}^2}{2}-2(-2\left)\right]$ $\begin{array}{c}=-[2-4-\frac{25}{2}+10]+[\frac{25}{2}+10-2+4]\\ =-2+4+\frac{25}{2}-10+\frac{25}{2}+10-2+4\\ =29\end{array}$

Q6. By using the properties of definite integrals, evaluate the integral. ${\int }_2^8|x-5|dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_2^6|x-5|dx\\ \text{It can be seen that}(x-5)\le 0\text{on}[2,5]\text{and}(x-5)\ge 0\text{on}[5,8]\end{array}$ $I={\int }_2^s-(x-5)dx+{\int }_2^s(x-5)dx\left({\int }_a^{b}f\right(x)={\int }_a^{t}f(x)+{\int }_c^{b}f(x\left)\right)$ $\begin{array}{cc}& =-{[\frac{x^2}{2}-5x]}^5+{[\frac{x^2}{2}-5x]}_5^8\\ & =-[\frac{25}{2}-25-2+10]+[32-40-\frac{25}{2}+25]\\ & =9\end{array}$

Q7. By using the properties of definite integrals, evaluate the integral. ${\int }_0^{1}x(1-x{)}^{n}dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_0^{1}x(1-x{)}^{n}dx\\ \therefore I={\int }_0^1(1-x)(1-(1-x){)}^{n}dx\end{array}$ $\begin{array}{c}={\int }_0^1(1-x)(x{)}^{n}dx\\ ={\int }_0^1(x^n-x^{n+1})dx\end{array}$ $={[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}]}_0^1\left({\int }_1^{0}f\right(x)dx={\int }_1^{0}f(a-x\left)dx\right)$ $\begin{array}{cc}& =[\frac{1}{n+1}-\frac{1}{n+2}]\\ & =\frac{(n+2)-(n+1)}{(n+1)(n+2)}\\ & =\frac{1}{(n+1)(n+2)}\end{array}$

Q8. By using the properties of definite integrals, evaluate the integral. ${\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx$

Answer. $I={\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx.....\left(1\right)$ $\therefore I={\int }_0^{\frac{\pi }{4}}\log [1+\tan (\frac{\pi }{4}-x)]dx\left({\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right)$ $\begin{array}{c}\Rightarrow I={\int }_0^{\frac{\pi }{4}}\log \{1+\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x}\}dx\\ \Rightarrow I={\int }_0^{\frac{\pi }{4}}\log \{1+\frac{1-\tan x}{1+\tan }\}dx\end{array}\}dx$ $\begin{array}{c}\Rightarrow I={\int }_0^{\frac{\pi }{4}}\log \frac{2}{(1+\tan x)}dx\\ \Rightarrow I={\int }_0^{\frac{\pi }{4}}\log 2dx-{\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx\end{array}$ $\Rightarrow I={\int }_0^{\frac{\pi }{4}}\log 2dx-I$ $\begin{array}{c}\Rightarrow 2I=[x\log 2{]}_0^{\frac{\pi }{4}}\\ \Rightarrow 2I=\frac{\pi }{4}\log 2\\ \Rightarrow I=\frac{\pi }{8}\log 2\end{array}$

Q9. By using the properties of definite integrals, evaluate the integral. ${\int }_0^{2}x\sqrt{2-x}dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_0^{2}x\sqrt{2-x}dx\\ I={\int }_0^2(2-x)\sqrt{x}dx\end{array}\left({\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right)$ $\begin{array}{c}={\int }_0^2\{2x^{\frac{1}{2}}-x^{\frac{3}{2}}\}dx\\ ={[2\left(\frac{x^{\frac{3}{2}}}{2}\right)-\frac{x^{\frac{5}{2}}}{2}]}_0^2\end{array}$ $\begin{array}{cc}& ={[\frac{4}{3}{x}^{\frac{3}{2}}-\frac{2}{5}{x}^{\frac{5}{2}}]}_0^2\\ & =\frac{4}{3}(2{)}^{\frac{3}{2}}-\frac{2}{5}(2{)}^{\frac{5}{2}}\end{array}$ $\begin{array}{cc}& =\frac{4\times 2\sqrt{2}}{3}-\frac{2}{5}\times 4\sqrt{2}\\ & =\frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5}\\ & =\frac{40\sqrt{2}-24\sqrt{2}}{15}\\ & =\frac{16\sqrt{2}}{15}\end{array}$

Q10. By using the properties of definite integrals, evaluate the integral. ${\int }_0^{\frac{\pi }{2}}(2\log \sin x-\log \sin 2x)dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_0^{\frac{\pi }{2}}(2\log \sin x-\log \sin 2x)dx\\ \Rightarrow I={\int }_0^{\frac{x}{2}}\{2\log \sin x-\log (2\sin x\cos x\left)\right\}dx\end{array}$ $\begin{array}{c}\Rightarrow I={\int }_0^{\frac{\pi }{2}}\{2\log \sin x-\log \sin x-\log \cos x-\log 2\}dx\\ \Rightarrow I={\int }_0^{\frac{\pi }{2}}\{\log \sin x-\log \cos x-\log 2\}dx.....\left(1\right)\end{array}$ It is known that, $\left({\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right)$ $\begin{array}{cc}\Rightarrow I={\int }_0^{\frac{\pi }{2}}\{\log \cos x-\log \sin x-\log 2\}dx& \dots \left(2\right)\\ \text{Adding}\left(1\right)\text{and}\left(2\right),\text{we obtain}& \end{array}$ $\begin{array}{c}2I={\int }_0^{\frac{\pi }{2}}(-\log 2-\log 2)dx\\ \Rightarrow 2I=-2\log 2{\int }_0^{\pi }1dx\\ \Rightarrow I=-\log 2\left[\frac{\pi }{2}\right]\end{array}$ $\begin{array}{c}\Rightarrow I=\frac{\pi }{2}\left[\log \frac{1}{2}\right]\\ \Rightarrow I=\frac{\pi }{2}\log \frac{1}{2}\end{array}$

Q11. By using the properties of definite integrals, evaluate the integral. ${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\sin }^{2}xdx$

Answer. $\begin{array}{c}\text{Let}I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\sin }^{2}xdx\\ \text{As}{\sin }^2(-x)=\left(\sin \right(-x){)}^2=(-\sin x{)}^2={\sin }^{2}x,\text{therefore,}{\sin }^{2}x\text{is an even function.}\end{array}$ It is known that if f(x) is an even function, then ${\int }_a^{-a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$ $\begin{array}{cc}I& =2{\int }_0^{\frac{\pi }{2}}{\sin }^{2}xdx\\ & =2{\int }_0^{\frac{\pi }{2}}\frac{1-\cos 2x}{2}dx\\ & ={\int }_0^{\frac{\pi }{2}}(1-\cos 2x)dx\\ & ={[x-\frac{\sin 2x}{2}]}_0^{\frac{\pi }{2}}\\ & =\frac{\pi }{2}\end{array}$

Q12. By using the properties of definite integrals, evaluate the integral. ${\int }_0^{\pi }\frac{xdx}{1+\sin x}$

Answer. $I={\int }_0^{\pi }\frac{xdx}{1+\sin x}.....\left(1\right)$ $\Rightarrow I={\int }_0^{\pi }\frac{(\pi -x)}{1+\sin (\pi -x)}dx\left({\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right)$ $\begin{array}{c}\Rightarrow I={\int }_0^{\pi }\frac{(\pi -x)}{1+\sin x}dx...\left(2\right)\\ \text{Adding}\left(1\right)\text{and}\left(2\right),\text{we obtain}\end{array}$ $\begin{array}{c}2I={\int }_0^{=}\frac{\pi }{1+\sin x}dx\\ \Rightarrow 2I=\pi {\int }_0^{\pi }\frac{(1-\sin x)}{(1+\sin x)(1-\sin x)}dx\\ \Rightarrow 2I=\pi {\int }_0^{=}\frac{1-\sin x}{{\cos }^{2}x}dx\end{array}$ $\begin{array}{c}\Rightarrow 2I=\pi [\tan x-\sec x{]}_0^{\pi }\\ \Rightarrow 2I=\pi \left[2\right]\\ \Rightarrow I=\pi \end{array}$

Q13. By using the properties of definite integrals, evaluate the integral. ${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{\sin }^{7}xdx$

Answer. $\text{Let I}={\int }_{\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^{7}xdx....\left(1\right)$ As ${\sin }^7(-x)=\left(\sin \right(-x){)}^7=(-\sin x{)}^7=-{\sin }^{7}x,\text{therefore,}{\sin }^{2}x$ is an odd function. $\begin{array}{c}\text{It is known that, if}f\left(x\right)\text{is an odd function, then}{\int }_{-a}^{a}f\left(x\right)dx=0\\ \therefore I={\int }_{\frac{\pi }{2}}^{\frac{\pi }{2}}{\sin }^{7}xdx=0\end{array}$

Q14. By using the properties of definite integrals, evaluate the integral. ${\int }_0^{2\pi }{\cos }^{5}xdx$

Answer. $\begin{array}{cc}\text{Let}I={\int }_0^{2\pi }{\cos }^{5}xdx& \dots \left(1\right)\\ {\cos }^5(2\pi -x)={\cos }^{5}x& \end{array}$ It is known that, $\begin{array}{cc}{\int }_0^{2a}f\left(x\right)dx& =2{\int }_1^{a}f\left(x\right)dx,\text{if}f(2a-x)=f\left(x\right)\\ & =0\text{if}f(2a-x)=-f\left(x\right)\end{array}$ $\therefore I=2{\int }_0^{\pi }{\cos }^{5}xdx$ $\Rightarrow I=2\left(0\right)=0\left[{\cos }^5\right(\pi -x)=-{\cos }^{5}x]$

Q15. By using the properties of definite integrals, evaluate the integral. ${\int }_0^{\frac{\pi }{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$

Answer. $\text{Let I}={\int }_0^{\frac{\pi }{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx....\left(1\right)$ $\Rightarrow I={\int }_0^{\pi }\frac{\sin (\frac{\pi }{2}-x)-\cos (\frac{\pi }{2}-x)}{1+\sin (\frac{\pi }{2}-x)\cos (\frac{\pi }{2}-x)}dx\left({\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right)$ $\Rightarrow I={\int }_0^{\frac{\pi }{2}}\frac{\cos x-\sin x}{1+\sin x\cos x}dx....\left(2\right)$ $\begin{array}{c}\text{Adding}\left(1\right)\text{and}\left(2\right),\text{we obtain}\\ 2I={\int }_b^{\frac{\pi }{2}}\frac{0}{1+\sin x\cos x}dx\\ \Rightarrow I=0\end{array}$

Q16. By using the properties of definite integrals, evaluate the integral. ${\int }_0^{\pi }\log (1+\cos x)dx$

Answer. $I={\int }_0^{\pi }\log (1+\cos x)dx.....\left(1\right)$ $I={\int }_0^{\pi }\log (1+\cos x)\left({\int }_1^{0}f\right(x)dx={\int }_1^{0}f(a-x\left)dx\right)$ $\Rightarrow I={\int }_1^{\pi }\log (1-\cos x)dx....\left(2\right)$ $\begin{array}{c}\text{Adding}\left(1\right)\text{and}\left(2\right),\text{we obtain}\\ 2I={\int }_0^{\pi }\left\{\log \right(1+\cos x)+\log (1-\cos x\left)\right\}dx\\ \Rightarrow 2I={\int }_0^{\pi }\log (1-{\cos }^{2}x)dx\\ \Rightarrow 2I={\int }_0^{\pi }\log {\sin }^{2}xdx\\ \Rightarrow 2I=2{\int }_0^{\pi }\log \sin xdx\end{array}$ $\begin{array}{c}\Rightarrow I={\int }_0^{\pi }\log \sin xdx\\ \sin (\pi -x)=\sin x.....\left(3\right)\end{array}$ $\begin{array}{c}\therefore I=2{\int }_0^{\frac{\pi }{2}}\log \sin xdx...\left(4\right)\\ \Rightarrow I=2{\int }_0^x\log \sin (\frac{\pi }{2}-x)dx=2{\int }_0^{\frac{\pi }{2}}\log \cos xdx....\left(5\right)\end{array}$ $\begin{array}{c}\text{Adding}\left(4\right)\text{and}\left(5\right),\text{we obtain}\\ 2I=2{\int }_0^{\frac{\pi }{2}}(\log \sin x+\log \cos x)dx\end{array}$ $\Rightarrow I={\int }_0^{\frac{x}{2}}(\log \sin x+\log \cos x+\log 2-\log 2)dx$ $\begin{array}{c}\Rightarrow I={\int }_0^{\frac{\pi }{2}}(\log 2\sin x\cos x-\log 2)dx\\ \Rightarrow I={\int }_0^{\frac{\pi }{2}}\log \sin 2xdx-{\int }_0^{\frac{\pi }{2}}\log 2dx\end{array}$ $\begin{array}{c}\text{Let}2x=t2dx=dt\\ \text{When}x=0,t=0\text{and when}x=\frac{\pi }{2},\pi =\\ \therefore I=\frac{1\pi }{2}{\int }_0^{\pi }\log \sin tdt-\frac{\log 2}{2}\log 2\end{array}$ $\begin{array}{c}\Rightarrow I=\frac{1\pi }{2}I-\frac{1}{2}\log 2\\ \Rightarrow \frac{I}{2}=-\frac{\pi }{2}\log 2\\ \Rightarrow I=-\pi \log 2\end{array}$

Q17. By using the properties of definite integrals, evaluate the integral. ${\int }_0^a\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx$

Answer. Let $I={\int }_0^0\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx....\left(1\right)$ It is known that, $\left({\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right)$ $I={\int }_0^a\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx.....\left(2\right)$ $\begin{array}{c}\text{Adding}\left(1\right)\text{and}\left(2\right),\text{we obtain}\\ 2I={\int }_0^a\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx\\ \Rightarrow 2I={\int }_0^{a}1dx\end{array}$ $\begin{array}{c}\Rightarrow 2I=[x{]}_0^a\\ \Rightarrow 2I=a\\ \Rightarrow I=\frac{a}{2}\end{array}$

Q18. By using the properties of definite integrals, evaluate the integral. ${\int }_0^4|x-1|dx$

Answer. $I={\int }_0^4|x-1|dx$ It can be seen that, (x − 1) ≤ 0 when 0 ≤ x ≤ 1 and (x − 1) ≥ 0 when 1 ≤ x ≤ 4 $I={\int }_0^4|x-1|dx+{\int }^4|x-1|dx\left({\int }_a^{b}f\right(x)={\int }_a^{c}f(x)+{\int }_t^{b}f(x\left)\right)$ $={\int }_0^1-(x-1)dx+{\int }_0^1(x-1)dx$ $={[x-\frac{x^2}{2}]}_0^1+{[\frac{x^2}{2}-x]}_1^4$ $=1-\frac{1}{2}+\frac{(4{)}^2}{2}-4-\frac{1}{2}+1$ $\begin{array}{c}=1-\frac{1}{2}+8-4-\frac{1}{2}+1\\ =5\end{array}$

Q19. By using the properties of definite integrals, evaluate the integral. Show that ${\int }_0^{a}f\left(x\right)g\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$, if f and g are defined as f(x) = f(a-x) and g(x) + g(a-x) = 4

Answer. $I={\int }_0^{a}f\left(x\right)g\left(x\right)dx....\left(1\right)$ $\Rightarrow I={\int }_1^{0}f(a-x)g(a-x)dx\left({\int }_1^{0}f\right(x)dx={\int }_0^{0}f(a-x\left)dx\right)$ $\begin{array}{c}\Rightarrow I={\int }_0^{a}f\left(x\right)g(a-x)dx.....\left(2\right)\\ \text{Adding}\left(1\right)\text{and}\left(2\right),\text{we obtain}\end{array}$ $\begin{array}{c}2I={\int }_0^a\left\{f\right(x\left)g\right(x)+f(x\left)g\right(a-x\left)\right\}dx\\ \Rightarrow 2I={\int }_0^{a}f\left(x\right)\left\{g\right(x)+g(a-x\left)\right\}dx\end{array}$ $\Rightarrow 2I={\int }_0^{\pi }f\left(x\right)\times 4dx\left[g\right(x)+g(a-x)=4]$ $\Rightarrow I=2{\int }_0^{a}f\left(x\right)dx$

Q20. Choose the correct answer The value of ${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}(x^3+x\cos x+{\tan }^{5}x+1)dx$is (A) 0 (B) 2 (C) π (D) 1

Answer. $\begin{array}{c}\text{Let}I={\int }_{\frac{\pi }{2}}^{\frac{\pi }{2}}(x^3+x\cos x+{\tan }^{5}x+1)dx\\ \Rightarrow I={\int }_{\frac{\pi }{2}}^{\frac{\pi }{2}}{x}^{3}dx+{\int }_{\frac{\pi }{2}}^{\frac{\pi }{2}}\cos x+{\int }_2^{\frac{\pi }{2}}{\tan }^{5}xdx+{\int }_2^{\frac{\pi }{2}}1\cdot dx\end{array}$ It is known that if f(x) is an even function, then ${\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx$ $\begin{array}{cc}I& =0+0+0+2{\int }_0^{\frac{\pi }{2}}1\cdot dx\\ & =2[x{]}_0^{\frac{\pi }{2}}\\ & =\frac{2\pi }{2}\\ \pi & =\end{array}$ Hence, the correct Answer is C.

Q21. Choose the correct answer The value of ${\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx$is $\begin{array}{cccc}\text{(A)}2& \text{(B)}\frac{3}{4}& \text{(C)}0& \text{(D)}-2\end{array}$

Answer. Let $I={\int }_0^2\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx$.....(1) $\Rightarrow I={\int }_0^{\pi }\log \left[\frac{4+3\sin (\frac{\pi }{2}-x)}{4+3\cos (\frac{\pi }{2}-x)}\right]dx\left({\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right)$ $\Rightarrow I={\int }_0^{\frac{\pi }{2}}\log \left(\frac{4+3\cos x}{4+3\sin x}\right)dx$.....(2) $\begin{array}{c}\text{Adding}\left(1\right)\text{and}\left(2\right),\text{we obtain}\\ 2I={\int }_0^{\frac{\pi }{2}}\{\log \left(\frac{4+3\sin x}{4+3\cos x}\right)+\log \left(\frac{4+3\cos x}{4+3\sin x}\right)\}dx\\ \Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\log (\frac{4+3\sin x}{4+3\cos x}\times \frac{4+3\cos x}{4+3\sin x})dx\end{array}$ $\begin{array}{c}\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\log 1dx\\ \Rightarrow 2I={\int }_0^{\frac{\pi }{2}}0dx\\ \Rightarrow I=0\end{array}$ 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 7.9
• Exercise 7.10