The value of \int\limits_ - \pi ^\pi \cos ^2 \over 1 + a^xdx,\,\,a \gt 0, is

Definite Integration PYQ — JEE Main 2005 · Shift 0 · Q71 with Solution

View written solutionFree

Correct answer: B

We need to evaluate $I=\int_{-\pi}^{\pi}\frac{\cos^2 x}{1+a^x}\,dx,\qquad a>0.$
Use the standard substitution $x\mapsto -x$ . Let $I=\int_{-\pi}^{\pi}\frac{\cos^2 x}{1+a^x}\,dx.$ Replacing $x$ by $-x$ , $I=\int_{-\pi}^{\pi}\frac{\cos^2 x}{1+a^{-x}}\,dx,$ because $\cos^2(-x)=\cos^2 x$ .
Simplify the new denominator term: $\frac{1}{1+a^{-x}}=\frac{a^x}{1+a^x}.$ So, $I=\int_{-\pi}^{\pi}\frac{a^x\cos^2 x}{1+a^x}\,dx.$
Add the two expressions for $I$ : $2I=\int_{-\pi}^{\pi}\left(\frac{\cos^2 x}{1+a^x}+\frac{a^x\cos^2 x}{1+a^x}\right)dx.$ Factor out $\cos^2 x$ :

\int_{-\pi}^{\pi}\cos^2 x\,dx.$$ Hence, $$I=\frac12\int_{-\pi}^{\pi}\cos^2 x\,dx.$$ 5. Now evaluate $\int_{-\pi}^{\pi}\cos^2 x\,dx$. Using $$\cos^2 x=\frac{1+\cos 2x}{2},$$ we get $$\int_{-\pi}^{\pi}\cos^2 x\,dx =\int_{-\pi}^{\pi}\frac{1+\cos 2x}{2}\,dx =\frac12\int_{-\pi}^{\pi}1\,dx+\frac12\int_{-\pi}^{\pi}\cos 2x\,dx.$$ Now, $$\frac12\int_{-\pi}^{\pi}1\,dx=\frac12(2\pi)=\pi,$$ and $$\int_{-\pi}^{\pi}\cos 2x\,dx=0.$$ Therefore, $$\int_{-\pi}^{\pi}\cos^2 x\,dx=\pi.$$ So, $$I=\frac12\cdot \pi=\frac{\pi}{2}.$$ 6. Hence the correct option is $$\boxed{\frac{\pi}{2}}.$$ So, option **B** is correct.

View written solutionFree

Correct answer: B

We need to evaluate $I=\int_{-\pi}^{\pi}\frac{\cos^2 x}{1+a^x}\,dx,\qquad a>0.$
Use the standard substitution $x\mapsto -x$ . Let $I=\int_{-\pi}^{\pi}\frac{\cos^2 x}{1+a^x}\,dx.$ Replacing $x$ by $-x$ , $I=\int_{-\pi}^{\pi}\frac{\cos^2 x}{1+a^{-x}}\,dx,$ because $\cos^2(-x)=\cos^2 x$ .
Simplify the new denominator term: $\frac{1}{1+a^{-x}}=\frac{a^x}{1+a^x}.$ So, $I=\int_{-\pi}^{\pi}\frac{a^x\cos^2 x}{1+a^x}\,dx.$
Add the two expressions for $I$ : $2I=\int_{-\pi}^{\pi}\left(\frac{\cos^2 x}{1+a^x}+\frac{a^x\cos^2 x}{1+a^x}\right)dx.$ Factor out $\cos^2 x$ :

\int_{-\pi}^{\pi}\cos^2 x\,dx.$$ Hence, $$I=\frac12\int_{-\pi}^{\pi}\cos^2 x\,dx.$$ 5. Now evaluate $\int_{-\pi}^{\pi}\cos^2 x\,dx$. Using $$\cos^2 x=\frac{1+\cos 2x}{2},$$ we get $$\int_{-\pi}^{\pi}\cos^2 x\,dx =\int_{-\pi}^{\pi}\frac{1+\cos 2x}{2}\,dx =\frac12\int_{-\pi}^{\pi}1\,dx+\frac12\int_{-\pi}^{\pi}\cos 2x\,dx.$$ Now, $$\frac12\int_{-\pi}^{\pi}1\,dx=\frac12(2\pi)=\pi,$$ and $$\int_{-\pi}^{\pi}\cos 2x\,dx=0.$$ Therefore, $$\int_{-\pi}^{\pi}\cos^2 x\,dx=\pi.$$ So, $$I=\frac12\cdot \pi=\frac{\pi}{2}.$$ 6. Hence the correct option is $$\boxed{\frac{\pi}{2}}.$$ So, option **B** is correct.