Question

Let $\int \frac{1-7co{s}^{2}x}{si{n}^{7}co{s}^{2}x}dx=\frac{g\left(x\right)}{si{n}^{7}x}+c$, then the value of $g^{\prime }\left(0\right)+g"\left(\frac{\pi }{4}\right)$ is

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

The correct option is C 5

Let $I=\int \frac{1-7co{s}^{2}x}{si{n}^{7}xco{s}^{2}x}dx$
$=\int \frac{dx}{si{n}^{7}xco{s}^{2}x}-\int \frac{7}{si{n}^{7}x}dx$
$=\int \frac{1}{si{n}^{7}x}.se{c}^{2}xdx-\int \frac{7}{si{n}^{7}x}dx$
$=\frac{1}{si{n}^{7}x}.tanx-\int \frac{-7cosx}{si{n}^{8}x}.tanxdx-\int \frac{7}{si{n}^{7}x}dx+c$
$=\frac{tanx}{si{n}^{7}x}+\int \frac{7}{si{n}^{7}x}dx-\int \frac{7}{si{n}^{7}x}+c$
$=\frac{tanx}{si{n}^{7}x}+c$
$g\left(x\right)=tanx,g^{\prime }\left(x\right)=se{c}^{2}xandg"\left(x\right)=2se{c}^{2}xtanx$
$\therefore g^{\prime }\left(0\right)+g"\left(\frac{\pi }{4}\right)=1+2(\sqrt{2}{)}^2=5$