Question

integrate using substitution tanx sec^2x dx

Answer 1

Answer:

$\frac{\tan^2(x)}{2}+C$

Step-by-step explanation:

Since I know $(\tan(x))'=\sec^2(x)$, then I will let $u=\tan(x)$ which will give me $du=\sec^2(x) dx$.

$\int \tan(x)\sec^2(x) dx$

$\int u du$ I replaced tan(x) with u and sec^2(x) dx with du.

$\int u^1 du$

Use the integration power rule:

$\frac{u^{1+1}}{1+1}+C$

$\frac{u^2}{2}+C$

Now put this back in terms of x with u=tan(x):

$\frac{\tan^2(x)}{2}+C$

Check:

To check we will find the derivative of our antiderivative and see it it gives us the integrand.

$\frac{d}{dx}(\frac{\tan^2(x)}{2}+C)$

By sum rule for derivatives:

$\frac{d}{dx}(\frac{\tan^2(x)}{2})+\frac{d}{dx}(C)$

Constant multiple rule will be applied to the first term while the constant rule will be applied to the second term:

$\frac{1}{2} \frac{d}{dx}(\tan^2(x))+0$

Power rule+Chain rule:

$\frac{1}{2}\cdot 2\tan(x) \cdot sec^2(x)$

Simplify the 1/2 times 2 part:

$\tan(x)\sec^2(x)$

This is what our integrand was so we have confirmed are integration was correct.