1. ## Integral

Hi,

Can you help me to calculate the following integral: $\int_0^{\frac{\pi}{4}} tan^3(t)+tan(t) dt= \int_0^{\frac{\pi}{4}}tan(t)(1+tan^2(t))dt \ = \ \int_0^{\frac{\pi}{4}}tan(t)tan'(t) dt=???$

2. Originally Posted by lehder
Hi,

Can you help me to calculate the following integral: $\int_0^{\frac{\pi}{4}} tan^3(t)+tan(t) dt= \int_0^{\frac{\pi}{4}}tan(t)(1+tan^2(t))dt \ = \ \int_0^{\frac{\pi}{4}}tan(t)tan'(t) dt=???$
Derivative of $(\tan(x))^2$ is...

3. Hello, lehder!

You're almost there . . .

$\int_0^{\frac{\pi}{4}} \left(\tan^3t+\tan t\right)\,dt$

We have: . $\int \tan x\left(\tan^2t + 1\right)\,dt \;=\;\int\tan x\,\sec^2\!x\,dx$

Let: $u \,=\,\tan x \quad\Rightarrow\quad du \,=\,\sec^2\!x\,dx$

Substitute: . $\int u\,du \;=\;\tfrac{1}{2}u^2+C$

Back-substitute: . $\tfrac{1}{2}\tan^2\!x + C$

Evaluate: . $\tfrac{1}{2}\tan^2\!x\,\bigg]^{\frac{\pi}{4}}_0 \;=\;\tfrac{1}{2}\tan^2\!\tfrac{\pi}{4} - \tfrac{1}{2}\tan^2\!0 \;=\;\tfrac{1}{2}(1^2) - \tfrac{1}{2}(0^2) \;=\;\frac{1}{2}$