$\int \frac{\sin x \cos^{3} x}{1 + \cos^{3} x} d x =$

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

$- \frac{1}{2} (1 + \cos^{2} x) - \frac{1}{2} \log (1 + \cos^{2} x) + c$

$\frac{1}{2} (1 + \cos^{2} x) - \frac{1}{2} \log (1 + \cos^{2} x) + c$

$\frac{- 1}{2} (1 + \cos^{2} x) + \frac{1}{2} \log (1 + \cos^{2} x) + c$

$\frac{1}{2} (1 + \cos^{2} x) + \frac{1}{2} \log (1 + \cos^{2} x) + c$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\int \frac{\sin x \cos^{3} x}{1 + \cos^{3} x} d x =$

$= \int \frac{t^{3}}{1 + t^{2}} (- d t) p u t \cos x = t \Rightarrow - \sin x d x = d t = - \int [t - \frac{t}{1 + t^{2}}] d t = \frac{1}{2} \int (\frac{2 t}{1 + t^{2}} - t) d t = \frac{1}{2} \log |1 + t^{2}| - \frac{t^{2}}{2} + C = \frac{1}{2} \log (1 + \cos^{2} x) - \frac{\cos^{2} x}{2} + C = \frac{1}{2} \log (1 + \cos^{2} x) - \frac{1}{2} (1 + \cos^{2} x) + (C + \frac{1}{2}) = \frac{1}{2} \log (1 + \cos^{2} x) - \frac{1}{2} (1 + \cos^{2} x) + K w h e r e C + \frac{1}{2} = K$