If f(x)=∫x2sin2x−sin2x(xtanx−1)(x2+1)2dx and f(0)=0 then f(π/4)=

∫(x2+1)sin2x−2xsin2x(x2+1)2dx =∫f(sin2xx2+1) f(x)=sin2xx2+1+cC=0 f(x)=sin2xx2+1 f(π4)=(1/2)2π216+1=1/2π216+1=8π2+16

If f(x)=∫x2sin2x−sin2x(xtanx−1)(x2+1)2dx and f(0)=0 then f(π/4)=

∫(x2+1)sin2x−2xsin2x(x2+1)2dx =∫f(sin2xx2+1) f(x)=sin2xx2+1+cC=0 f(x)=sin2xx2+1 f(π4)=(1/2)2π216+1=1/2π216+1=8π2+16

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$I f f (x) = \int \frac{x^{2} \sin 2 x - \sin 2 x (x \tan x - 1)}{{(x^{2} + 1)}^{2}} d x a n d f (0) = 0 t h e n f (π / 4) =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{8}{π^{2} + 16}$

$\frac{4}{π^{2} + 16}$

$\frac{1}{π^{2} + 16}$

$\frac{1}{π^{2} + 9}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\int \frac{(x^{2} + 1) \sin 2 x - 2 x \sin^{2} x}{{(x^{2} + 1)}^{2}} d x = \int f (\frac{\sin^{2} x}{x^{2} + 1}) \begin{array}{l} f (x) = \frac{\sin^{2} x}{x^{2} + 1} + c \\ C = 0 \end{array} f (x) = \frac{\sin^{2} x}{x^{2} + 1} f (\frac{π}{4}) = \frac{{(1 / \sqrt{2})}^{2}}{\frac{π^{2}}{16} + 1} = \frac{1 / 2}{\frac{π^{2}}{16} + 1} = \frac{8}{π^{2} + 16}$