Mathematics

Evaluate: $$\displaystyle\int \dfrac{\sin^3x}{( \cos^4x+ 3 \cos^2x+ 1)\tan^{-1} ( \sec x+ \cos x)}dx$$


ANSWER

$$\log_e | \tan^{-1} ( \sec x + \cos x) |+c$$


SOLUTION
we have to evaluate
$$I= \int\dfrac{sin^3\ x}{(cos^4x+3\  cos^2+1)tan^{-1}(sec\ x+ cos\ x)}dx$$

Let us assume $$tan^{-1}(sec\  x+ cos\ x)=t$$

then,
$$\Rightarrow \dfrac{1}{1+ (sec\ x+cos\ x)^2}.(sec\ x.tan\ x- sin\ x)dx=dt$$

$$\Rightarrow \dfrac{1}{1+ (\dfrac{1}{cos\ x}+cos\ x)^2}.(\dfrac{sin\ x}{cos^2\ x}- sin\ x)dx=dt$$

$$\Rightarrow \dfrac{cos^2\ x}{cos^2\ x+ (1+cos^2\ x)^2}.\dfrac{(sin\ x- sin\ xcos^2\ x)}{cos^2\ x}dx=dt$$

$$\Rightarrow \dfrac{sin\ x(1- cos^2x)}{cos^2x+1+ cos^4x+2cos^2x}dx=dt$$


$$\Rightarrow \dfrac{sin^3x}{ cos^4x+3cos^2x+ 1}dx=dt$$

$$I= \int\dfrac{sin^3x}{(cos^4x+3 cos^2+1)tan^{-1}(sec\ x+ cos\ x)}dx\Rightarrow\int\dfrac{dt}{t}$$

$$I = log_e\begin{vmatrix}t\end{vmatrix}+ c$$

$$I = log_e\begin{vmatrix}tan^{-1}(sec\ x +cos\ x)\end{vmatrix}+ C$$

$$\therefore \text{ Correct answer is option B.}$$ 
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Single Correct Medium Published on 17th 09, 2020
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