# unique's questions at Yahoo! Answers regarding integration

#### MarkFL

Staff member
Here are the questions:

Find the following integrals please show steps?

find the following integrals

a)integral -pi/2,pi/2(2t+cost)dt

b)integral x^3/(x^4+2)^2 dx

c)integral -2,2x(x^2+3)^2 dx

d)integral 0,pi/4 tan x sec^2 x dx
I have posted a link there to this topic so the OP can see my work.

#### MarkFL

Staff member
Hello unique,

We are given to evaluate:

a) $$\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}2t+\cos(t)\,dt$$

I would write this as the sum of two integrals:

$$\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}2t\,dt+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(t)\,dt$$

Now, by the odd and even rules, we may write:

$$\displaystyle I=0+2\int_{0}^{\frac{\pi}{2}}\cos(t)\,dt=2\left[\sin(t) \right]_{0}^{\frac{\pi}{2}}=2(1-0)=2$$

b) $$\displaystyle I=\int\frac{x^3}{\left(x^4+2 \right)^2}\,dx$$

Let:

$$\displaystyle u=x^4+2\,\therefore\,du=4x^3\,dx$$

and we have:

$$\displaystyle I=\frac{1}{4}\int u^{-2}\,du=-\frac{1}{4u}+C=-\frac{1}{4\left(x^4+2 \right)}+C$$

c) $$\displaystyle I=\int_{-2}^2 x(x^2+3)^2\,dx$$

By the odd-function rule, we have:

$$\displaystyle I=0$$

d) $$\displaystyle I=\int_0^{\frac{\pi}{4}} \tan(x)\sec^2(x)\,dx$$

Let:

$$\displaystyle u=\tan(x)\,\therefore\,du=\sec^2(x)\,dx$$

and we have:

$$\displaystyle I=\int_0^1 u\,du=\frac{1}{2}\left[u^2 \right]_0^1=\frac{1}{2}(1-0)=\frac{1}{2}$$