s i n 3 x ( c o s 4 x + 3 c o s 2 x + 1 ) t a n − 1 ( s e c x + c o s x ) d x = Options: A . &nbsptan−1(secx+cosx)+c B . &nbsplog|tan−1(secx+cosx)|+c C . &nbsp1(secx+cos2x)2+c D . &nbsplog|secx+cosx|+c

Answer: Option B : B Put t a n − 1 ( s e c x + c o s x ) = f ( x ) f 1 ( x ) = s i n 3 x c o s 4 x + 3 c o s 2 x + 1 ∴ ∫ f 1 ( x ) f ( x ) = l o g | f ( x ) | + c

sin3x(cos4x+3cos2x+1)tan−1(secx+cosx)dx=

Question

\frac{s i n^{3} x}{(c o s^{4} x + 3 c o s^{2} x + 1) t a n^{- 1} (s e c x + c o s x)} d x =

Options:

A . tan−1(secx+cosx)+c

B . log|tan−1(secx+cosx)|+c

C . 1(secx+cos2x)2+c

D . log|secx+cosx|+c

Answer: Option B
:
B
Put

t a n^{- 1} (s e c x + c o s x) = f (x) f^{1} (x) = \frac{s i n^{3} x}{c o s^{4} x + 3 c o s^{2} x + 1} ∴ \int \frac{f^{1} (x)}{f (x)} = l o g | f (x) | + c

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