Question

$\int \frac{\left(se{c}^{2}x-7\right)}{\left({\sin }^{7}x\right)}dx=$

• A. $\left[\frac{\tan x}{{\sin }^{7}x}\right]+c$
• B. $\left[\frac{\cos x}{{\sin }^{7}x}\right]+c$
• C. $\left[\frac{\sin x}{{\cos }^{7}x}\right]+c$
• D. $\left[\frac{\sin x}{{\tan }^{7}x}\right]+c$

Answer 1

The correct option is A

$\left[\frac{\tan x}{{\sin }^{7}x}\right]+c$

Explanation for the correct option:

Evaluate $\int \frac{\left(se{c}^{2}x-7\right)}{\left({\sin }^{7}x\right)}dx$

Consider the given equation as

$$\begin{gathered} I=\int \frac{\left(se{c}^{2}x-7\right)}{\left({\sin }^{7}x\right)}dx \\ I=\int \cos e{c}^{7}x.se{c}^{2}xdx-\int 7\cos e{c}^{7}xdx \end{gathered}$$

According to the integration by parts

$$\begin{gathered} I=\cos e{c}^{7}x.\tan x-\int 7\cos e{c}^{6}x\left(-\cos ecx.cotx\right)\tan xdx-7\int \cos e{c}^{7}xdx+c \\ I=\cos e{c}^{7}x.\tan x+\int 7\cos e{c}^{7}xdx-7\int \cos e{c}^{7}xdx+c \\ I=\cos e{c}^{7}x.\tan x+c \\ I=\frac{\tan x}{{\sin }^{7}x}+c \end{gathered}$$

Hence, the correct answer is option A.