trig integrals

**skabani** · September 16th 2008, 02:28 PM

i have a calc test coming up and im not sure how to do the following, if you could show me how, it would be greatly appreciated...thanks!

1. integral (where b= pi/2, a = 0) sin^2x cos^2 x dx

2. integral (where b= pi/4, a = 0) sec^4 x tan^4 x dx

3. integral tan^6(ay) dy

**Soroban** · September 16th 2008, 03:36 PM

Hello, skabani!

Here are the first two.
. . I'll let someone else explain the third one.

$1)\;\;\int^{\frac{\pi}{2}}_0 \sin^2\!x\cos^2\!x\,dx$

We need two identities:
. . $2\sin\theta\cos\theta \:=\:\sin2\theta\qquad\qquad \sin^2\!\theta \:=\:\frac{1 - \cos2\theta}{2}$

We have: . $\sin^2\!x\cos^2\!x \;=\;\frac{1}{4}\left(4\sin^2\!x\cos^2\!x\right) \;=\;\frac{1}{4}(2\sin x\cos x)^2 \;=\;\frac{1}{4}\sin^22x$

The integral becomes: . $\frac{1}{4}\int^{\frac{\pi}{2}}_0\sin^22x\,dx \;=\; \frac{1}{4}\int^{\frac{\pi}{2}}_0 \frac{1-\cos4x}{2}\,dx$

And we have: . $\frac{1}{8}\int^{\frac{\pi}{2}}_0(1 - \cos4x)\,dx\quad\hdots$ . Got it?

$2)\;\;\int^{\frac{\pi}{4}}_0 \sec^4\!x\tan^4\!x\,dx$

We have: . $\sec^2\!x\cdot\sec^2\!x\cdot\tan^4\!x \;=\;\sec^2\!x\cdot(\tan^2\!x + 1)\cdot\tan^4\!x<br /> \;=\;\sec^2\!x\cdot(\tan^6\!x+\tan^4\!x)$

The integral becomes: . $\int^{\frac{\pi}{4}}_0 (\tan^6\!x + \tan^4\!x)\,\sec^2\!x\,dx$

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

Substitute: . $\int^b_a(u^6 + u^4)\,du \quad\hdots$ . Okay?

**skabani** · September 16th 2008, 03:44 PM

great thanks so much!

i hope someone can help me out with the third one, before tonight, for my exam tomorrow...

**Krizalid** · September 16th 2008, 04:06 PM

For the last one, first put $z=\alpha y,$ then just worry about $\int{\tan ^{6}z\,dz}.$ Now, note that,

$\tan ^{6}z=\tan ^{4}z\tan ^{2}z=\tan ^{4}z\left( \sec ^{2}z-1 \right)=\tan ^{4}z\sec ^{2}z-\tan ^{4}z.$

Make it in the same fashion for $\tan ^{4}z.$

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