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Question: The value of \(\int_{}^{}{\frac{\cos^{3}x + \cos^{5}x}{\sin^{2}x + \sin^{4}x}dx}\) is...

The value of ∫cos⁡3x+cos⁡5xsin⁡2x+sin⁡4xdx\int_{}^{}{\frac{\cos^{3}x + \cos^{5}x}{\sin^{2}x + \sin^{4}x}dx} is

A

sin⁡x−6tan⁡−1(sin⁡x)+C\sin x - 6\tan^{- 1}\left( \sin x \right) + C

B

sin⁡x−2(sin⁡x)−1+C\sin x - 2\left( \sin x \right)^{- 1} + C

C

sin⁡x−2(sin⁡x)−1−6tan⁡−1(sin⁡x)+C\sin x - 2\left( \sin x \right)^{- 1} - 6\tan^{- 1}\left( \sin x \right) + C

D

sin⁡x−2(sin⁡x)−1+5tan⁡−1(sin⁡x)+C\sin x - 2\left( \sin x \right)^{- 1} + 5\tan^{- 1}\left( \sin x \right) + C

Answer

sin⁡x−2(sin⁡x)−1−6tan⁡−1(sin⁡x)+C\sin x - 2\left( \sin x \right)^{- 1} - 6\tan^{- 1}\left( \sin x \right) + C

Explanation

Solution

R(sin⁡x,cos⁡x)=cos⁡3x+cos⁡5xsin⁡2x+sin⁡4xR ( \sin x , \cos x ) = \frac { \cos ^ { 3 } x + \cos ^ { 5 } x } { \sin ^ { 2 } x + \sin ^ { 4 } x }⇒ R(sin⁡x,−cos⁡x)=−cos⁡3x−cos⁡5xsin⁡2x+sin⁡4x=−cos⁡3x+cos⁡5xsin⁡2x+sin⁡4xR ( \sin x , - \cos x ) = \frac { - \cos ^ { 3 } x - \cos ^ { 5 } x } { \sin ^ { 2 } x + \sin ^ { 4 } x } = - \frac { \cos ^ { 3 } x + \cos ^ { 5 } x } { \sin ^ { 2 } x + \sin ^ { 4 } x }

= −R(sin⁡x,cos⁡x)- R\left( \sin x,\cos x \right)

Therefore, we substitute sinx = t, so that cosxdx= dt.

∫cos⁡3x+cos⁡5xsin⁡2x+sin⁡4xdx=∫cos⁡x[(1−sin⁡2x)+(1−sin⁡2x)2]sin⁡2x+sin⁡4xdx\int_{}^{}{\frac{\cos^{3}x + \cos^{5}x}{\sin^{2}x + \sin^{4}x}dx = \int_{}^{}{\frac{\cos x\left\lbrack \left( 1 - \sin^{2}x \right) + \left( 1 - \sin^{2}x \right)^{2} \right\rbrack}{\sin^{2}x + \sin^{4}x}dx}}

∫(1−t2)(2−t2)t2+t4dt=∫(1+2t2−61+t2)dt\int_{}^{}{\frac{\left( 1 - t^{2} \right)\left( 2 - t^{2} \right)}{t^{2} + t^{4}}dt = \int_{}^{}{\left( 1 + \frac{2}{t^{2}} - \frac{6}{1 + t^{2}} \right)dt}}

=t−2t−6tan⁡−1t+Ct - \frac{2}{t} - 6\tan^{- 1}t + C = sin⁡x−2(sin⁡x)−1−6tan⁡−1(sin⁡x)+C\sin x - 2\left( \sin x \right)^{- 1} - 6\tan^{- 1}\left( \sin x \right) + C