# Evaluating integrals

• May 5th 2009, 11:23 AM
bosmith
Evaluating integrals
ò sin ^5 x dx
ò tan 3 x dx
Do I simply apply rule one for adding the the exponent and placing the same exponent in the denominator? Do I substitue -cos for sin? Do I substitute sec^2 x for tan x? Help please.
• May 5th 2009, 11:47 AM
redsoxfan325
Quote:

Originally Posted by bosmith
ò sin ^5 x dx
ò tan 3 x dx
Do I simply apply rule one for adding the the exponent and placing the same exponent in the denominator? Do I substitue -cos for sin? Do I substitute sec^2 x for tan x? Help please.

Note: I'll be using the fact that $\sin^2(x)=1-\cos^2(x)$ a lot without explicitly stating it.

$\int\sin^5(x)\,dx = \underbrace{\int\sin^3(x)\,dx}_{1st~integral}-\underbrace{\int\sin^3(x)\cos^2(x)\,dx}_{2nd~integ ral}$

First Integral: $\int\sin^3(x) = \int\sin(x)-\sin(x)\cos^2(x)\,dx$

Let $u=\cos(x)$ so $du = -\sin(x)\,dx$. We have: $-\int\,du +\int u^2\,du = -u+\frac{u^3}{3} = -\cos(x)+\frac{\cos^3(x)}{3}$

Second Integral: $\int\sin^3(x)\cos^2(x)\,dx = \int\sin(x)\cos^2(x)-\sin(x)\cos^4(x)\,dx$

Let $u=\cos(x)$ so $du = -\sin(x)\,dx$. We have:

$-\int u^2\,du +\int u^4\,du = -\frac{u^3}{3}+\frac{u^5}{5} = -\frac{\cos^3(x)}{3}+\frac{\cos^5(x)}{5}$

$-\cos(x)+\frac{\cos^3(x)}{3} - \left(-\frac{\cos^3(x)}{3}+\frac{\cos^5(x)}{5}\right) = \boxed{-\frac{1}{5}\cos^5(x) + \frac{2}{3}\cos^3(x)-\cos(x)}$
• May 5th 2009, 11:52 AM
redsoxfan325
Quote:

Originally Posted by bosmith
ò sin ^5 x dx
ò tan 3 x dx
Do I simply apply rule one for adding the the exponent and placing the same exponent in the denominator? Do I substitue -cos for sin? Do I substitute sec^2 x for tan x? Help please.

$\int\tan(3x)\,dx = \int\frac{\sin(3x)}{\cos(3x)}\,dx$

Let $u=\cos(3x)$ and $du=-3\sin(3x)\,dx$. We now have:

$-\frac{1}{3}\int\frac{1}{u}\,du = -\frac{1}{3}\ln|u| = \boxed{-\frac{1}{3}\ln|\cos(3x)|}$

Or did you mean $\int\tan^3(x)\,dx$ ?
• May 5th 2009, 12:27 PM
bosmith
redsoxfan325
Thanks for showing me that I was on the wrong track.
• May 5th 2009, 12:29 PM
HallsofIvy
Quote:

Originally Posted by bosmith
ò sin ^5 x dx
ò tan 3 x dx
Do I simply apply rule one for adding the the exponent and placing the same exponent in the denominator? Do I substitue -cos for sin? Do I substitute sec^2 x for tan x? Help please.

I'm not sure what you mean by "adding the exponent". If you were thinking of the power rule, no, because this is $x^n dx$, you cannot do that.

Because of the odd power in $sin^5 t$, you can factor out one "sin t" to use with the derivative: $\int sin^5 x dx= \int sin^4 x (sin x dx)$ $= (1- cos^2 x)^2 (sin x dx)$. Now use the substitution u= cos x, du= -sin x dx so the integral becomes $-\int (1- u^2)^2 du$. Multiply that out and you can apply the power rule to each part.

For $\int tan^3 x dx$ you can't "substitute sec^2 x for tan x"- they are not equal! It is the derivative of tan x that is equal to $sec^2 x$ and you don't have any derivative of tan x in this. I would write it in the form $\int \frac{sin^2 x}{cos^3 x} dx$. Again, that is an odd power of sine so we can factor out one to use with dx:
$\int \frac{sin^2 x}{cos^3 x}(sin x dx)$. Again, let u= cos(x) so that du= -sin x dx so the integral becomes $-\int \frac{1- cos^2 x}{cos^3 x} (cos x dx)= -\int \frac{1- u^2}{u^3}du$.
• May 5th 2009, 12:35 PM
bosmith
hallsofivy
Thank you. I am struggling trying to lean Calculus. I was on the wrong track (confused). I now have a better understanding of the substitution fule, usign u= cosx. I will keep trying. I guess I am just not that smart. I have to leanr to ask questions sooner or somehow try to comprehend faster. I have a test in a couple of hours. Too bad for me that nobody seems to think it beneficial just to show me how to do the problems. But I truly thank you for putting me on the right track.