$\int \sec ^4 x \cdot an ^4 x d x=\frac{ an ^m x}{m}+\frac{ an ^n x}{n}+c$ (where $c$ is constant of integration), then $\mathrm{m}+\mathrm{n}=$

Question

$\int \sec ^4 x \cdot 	an ^4 x d x=\frac{	an ^m x}{m}+\frac{	an ^n x}{n}+c$ (where $c$ is constant of integration), then $\mathrm{m}+\mathrm{n}=$

Aicharya · Accepted Answer

Let $I=\int \sec ^4 	an ^4 x d x$ Put $	an \mathrm{x}=\mathrm{t} \Rightarrow \sec ^2 \mathrm{x} d \mathrm{x}=\mathrm{dt}$ $\begin{aligned} & 	herefore \mathrm{I}=\int \sec ^2 \mathrm{x}\left(\sec ^2 \mathrm{x}ight)\left(	an ^4 \mathrm{x}ight) \mathrm{dx} \ & =\int\left(1+\mathrm{t}^2ight)(\mathrm{t})^4 \mathrm{dt}=\int\left(\mathrm{t}^4+\mathrm{t}^6ight) \mathrm{dt}=\frac{\mathrm{t}^5}{5}+\frac{\mathrm{t}^7}{7}+\mathrm{c}=\frac{	an ^5 \mathrm{x}}{5}+\frac{	an ^7 \mathrm{x}}{7}+\mathrm{c} \end{aligned}$

$\int \sec ^4 x \cdot \tan ^4 x d x=\frac{\tan ^m x}{m}+\frac{\tan ^n x}{n}+c$ (where $c$ is constant of…

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