Question

Solve this one !
standard :-12th
chapter:- definite integration pattern:- jee / advance

Question :- given in attachment .​

Answer 1

Answer is given in the pic.

Answer 2

Answer:

$I = \frac{ {\pi}^{2} }{8}$

Step-by-step explanation:

• We know that

$\int^{a}_{b}f(x)dx + \int^{f(a)}_{f(b)}f^{ - 1} (x)dx = af(a) - bf(b)$

• Clearly comparing the given integral with the above equation, we get :

$\int^{ \frac{\pi}{2} }_{0}arctan( \sin(x)) dx + \int^{ \frac{\pi}{4} }_{0}arcsin( \tan(x) )dx = \frac{\pi}{2} \times \frac{ \pi}{4} = \frac{ {\pi}^{2} }{8}$