Integration Drill 2

Calculus Level 3

$\large I_{n} = \int (\tan x)^n dx$

For $I_n$ as defined above, find $I_{49} + I_{47}$ .

\frac{(\tan x)^{47}}{47}

\frac{(\tan x)^{48}}{48}

\frac{(\sec x)^{96}}{96}

\frac{(\sec x)^{94}}{94}

1 solution

X X
Jul 13, 2018

$\space\space\space\space\displaystyle\int (\tan x)^{47}+(\tan x)^{49} dx$

$\displaystyle=\int (\tan x)^{47}(\tan^2 x+1) dx$

$\displaystyle=\int (\tan x)^{47}\sec^2 xdx$

$\displaystyle=\int u^{47}du\space\space\space\space\space\space\space\space\space\color{#3D99F6}\text{Put} \space u=\tan x,du=\sec^2xdx$

$\displaystyle=\frac{u^{48}}{48}+C$

$\displaystyle=\frac{(\tan x)^{48}}{48}+C$