please integrate : (tan x)^2/3 its urgent

shubham

Deepak Aggarwal

Put tan(x) = y³

So, sec²x dx = 3y² dy

So, integral becomes int 3y⁴ / 1 + y⁶ dy

This integral i suppose u can solve now.

akki ~~ unlucky forever ~~

$I=\int (tan\theta)^\frac{2}{3}d\theta\\ \\ \\ put\ tan\theta=\frac{1}{u^3}\\ \\ \\ I=\int \frac{-3\ du}{(u^6+1)}\\ \\ \\ I=-3 \int \frac{du}{(u^2+1)(u^2+\sqrt{3}u+1)(u^2-\sqrt{3}u+1)}\\ \\ \\ I=\frac{-\sqrt{3}}{2} \int \frac{1}{u(u^2+1)}(\frac{1}{u^2-\sqrt{3}u+1}-\frac{1}{u^2+\sqrt{3}u+1})du\\ \\ \\ I=\frac{1}{2} \int \frac{1}{u^2}(\frac{1}{u^2+1}-\frac{1}{u^2-\sqrt{3}u+1})du+\frac{1}{2} \int \frac{1}{u^2}(\frac{1}{u^2+1}-\frac{1}{u^2+\sqrt{3}u+1})du\\ \\ \\ I=\int \frac{du}{u^2}-\int \frac{du}{u^2+1}-\frac{1}{2} \int (\frac{1}{u^2(u^2-\sqrt{3}u+1)}+\frac{1}{u^2(u^2+\sqrt{3}u+1)})du\\ \\ \\ I=-\int \frac{du}{u^2+1}-\frac{1}{2}\int (\frac{\sqrt{3}u+2}{u^2+\sqrt{3}u+1})du-\frac{1}{2} \int( \frac{-\sqrt{3}u+2}{u^2-\sqrt{3}u+1})du\\ \\ \\ now\ each\ of\ the\ 3\ integrals\ r\ in\ their\ standard\ format\\ &\ hence\ can\ easily\ be\ solved.$

.

Sagar Saxena

hiiiii

!!!!!!

Ask & Discuss Questions with Community & Experts