Question

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

• A. ${\tan }^{-1}(\sec x+\cos x)+c$
• B. ${\log }_e|{\tan }^{-1}(\sec x+\cos x)|+c$
• C. $\frac{1}{(\sec x+\cos x{)}^2}+c$
• D. ${\tan }^{-1}(\cot x+\text{cosec}x)+c$

Answer 1

The correct option is B ${\log }_e|{\tan }^{-1}(\sec x+\cos x)|+c$

we have to evaluate

$I=\int \frac{si{n}^{3}x}{(co{s}^{4}x+3co{s}^2+1)ta{n}^{-1}(secx+cosx)}dx$

Let us assume $ta{n}^{-1}(secx+cosx)=t$

then,

$\Rightarrow \frac{1}{1+(secx+cosx{)}^2}.(secx.tanx-sinx)dx=dt$

$\Rightarrow \frac{1}{1+(\frac{1}{cosx}+cosx{)}^2}.(\frac{sinx}{co{s}^{2}x}-sinx)dx=dt$

$\Rightarrow \frac{co{s}^{2}x}{co{s}^{2}x+(1+co{s}^{2}x{)}^2}.\frac{(sinx-sinxco{s}^{2}x)}{co{s}^{2}x}dx=dt$

$\Rightarrow \frac{sinx(1-co{s}^{2}x)}{co{s}^{2}x+1+co{s}^{4}x+2co{s}^{2}x}dx=dt$

$\Rightarrow \frac{si{n}^{3}x}{co{s}^{4}x+3co{s}^{2}x+1}dx=dt$

$I=\int \frac{si{n}^{3}x}{(co{s}^{4}x+3co{s}^2+1)ta{n}^{-1}(secx+cosx)}dx\Rightarrow \int \frac{dt}{t}$

$I=lo{g}_e\left|\begin{array}{c}t\end{array}\right|+c$

$I=lo{g}_e\left|\begin{array}{c}ta{n}^{-1}(secx+cosx)\end{array}\right|+C$

$\therefore \text{Correct answer is option B.}$