Definite Integration
Integral Calculus-2
star_batch_jee_advanced_2025
Grade None
Question:
Let $u = \int_0^{\pi/4} \left(\frac{\cos x}{\sin x + \cos x}\right)^2 dx$ and $v = \int_0^{\pi/4} \left(\frac{\sin x + \cos x}{\cos x}\right)^2 dx$, then:
$v = 1 + \ln 2$
$u = \frac{1 + \ln 2}{4}$
$\frac{v}{u} = 6$
$\frac{u}{v} = \frac{1}{6}$
Step-by-Step Solution
Key Concept: Decomposing the integrand and using the double angle formula $\sin 2x = 2\sin x\cos x$ simplifies rational integrals in trigonometric form.
For $v = -\int_0^{\pi/4} \frac{1+\sin 2x}{\cos^2 x}dx$, expand as $\int_0^{\pi/4}(\sec^2 x + 2\tan x)dx = [\tan x + 2\ln(\sec x)]_0^{\pi/4} = 1 + \ln 2$. For $u = \int_0^{\pi/4} \frac{\cos^2 x}{1+\sin 2x}dx$, use $\sin 2x = \frac{2\tan x}{1+\tan^2 x}$ and substitute $t = \tan x$ to reduce to $u = \frac{1}{4}[\ln 2 + 1]$. Thus $\frac{v}{u} = 4$.
Correct Answer: 1,2