int_{frac{pi}{5}}^{frac{3 pi}{10}} frac{tan x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $I = \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{	an x}{	an x + \cot x} \, dx \quad \dots(1)$ Using the property $\int_{a}^{b} f(x) \, dx

Vedclass · Accepted Answer

$\frac{\pi}{20}$

Vedclass · Answer

(D) Let $I = \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{	an x}{	an x + \cot x} \, dx \quad \dots(1)$ Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,where $a = \frac{\pi}{5}$ and $b = \frac{3 \pi}{10}$,we have $a+b = \frac{\pi}{5} + \frac{3 \pi}{10} = \frac{2\pi + 3\pi}{10} = \frac{5\pi}{10} = \frac{\pi}{2}$. So,$I = \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{	an(\frac{\pi}{2}-x)}{	an(\frac{\pi}{2}-x) + \cot(\frac{\pi}{2}-x)} \, dx$. Since $	an(\frac{\pi}{2}-x) = \cot x$ and $\cot(\frac{\pi}{2}-x) = 	an x$,we get: $I = \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{\cot x}{\cot x + 	an x} \, dx \quad \dots(2)$ Adding equations $(1)$ and $(2)$: $2I = \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{	an x + \cot x}{	an x + \cot x} \, dx = \int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} 1 \, dx$ $2I = [x]_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} = \frac{3 \pi}{10} - \frac{\pi}{5} = \frac{3 \pi - 2 \pi}{10} = \frac{\pi}{10}$ $I = \frac{\pi}{20}$

$\int_{\frac{\pi}{5}}^{\frac{3 \pi}{10}} \frac{\tan x}{\tan x + \cot x} \, dx =$

Similar Questions

The value of $\int_{0}^{\pi} \log (1+\cos x) d x$ is

$\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=$

Evaluate the definite integral $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x$.

$\int_{0}^{\frac{\pi}{2}}\left(e^{\sin x}-e^{\cos x}\right) d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module