# Even powers of $$\sin ⁡x$$ and $$\cos ⁡x$$

LESSON 1
Determine
$\int \sin^4⁡x\:dx$
SOLUTION
• Rewrite as multiples of $$\sin^2⁡x$$
$$\int \sin^2⁡x \sin^2⁡x \:dx$$
• Use the identity
$$\displaystyle \sin^2⁡x={1-\cos ⁡2x\over2}$$
Therefore, we now have
$$\displaystyle \sin^4⁡x\:dx=\int \left({1\over 2}-{\cos⁡2x\over2}\right)\left({1\over2}-\cos{⁡2x\over2}\right)\:dx$$
$$\displaystyle =\int\left({1\over4}-\cos⁡{2x\over2}+\cos^2⁡{2x\over4}\right)\:dx$$
$$\displaystyle ={1\over4}\int(1-2 \cos⁡2x+\cos^2⁡2x)\: dx$$
$$\displaystyle ={1\over4} \left[\int1\: dx-2\int\cos⁡2x\:dx+\int\cos^2⁡2x\: dx\right]$$
$$\displaystyle ={1\over4} \left[x-2\left({1\over2} \sin⁡2x\right )+\int \cos^2⁡2x \: dx\right]$$

• To integrate $$\cos^2⁡2x$$ use the double angle formula
$$\cos⁡2x=2 \cos^2⁡x-1$$
$$⟹\cos⁡(2(2x))=2 \cos^2⁡(2x)-1$$
$$⟹\cos⁡4x=2 \cos^2⁡2x-1$$
$$\displaystyle ⟹\cos^2⁡2x={\cos⁡4x+1\over2}$$
$$\displaystyle ⟹\int \cos^2⁡2x \:dx={1\over2} \int (\cos⁡4x+1) \:dx$$
$$\displaystyle ={1\over2} \left[{1\over4} \sin⁡4x+x\right]$$
$$\displaystyle ={1\over8} \sin⁡4x+{x\over2}$$
• Resubstituting
$$\displaystyle \int \sin^4⁡x \:dx={1\over4} \left[x-\sin⁡2x+\left({\sin⁡4x\over8}+{x\over2}\right)\right]+c$$
$$\displaystyle ={x\over4}-{\sin⁡2x\over4}+{\sin⁡4x\over32}+{x\over8}+c$$
$$\displaystyle ={3x\over8}-{\sin⁡2x\over4}+{\sin⁡4x\over32}+c$$
$$\displaystyle ={3x\over8}-{\sin⁡2x\over4}+{\sin⁡4x\over32}+c$$

In the previous example, we saw how to integrate even multiples of cosine but let’s look at it once again.
LESSON 2
Determine
$\int \cos^2⁡x \:dx$
SOLUTION
$$\int \cos^2⁡x \:dx$$
• Use the identity
$$\cos⁡2x=2 \cos^2⁡x-1$$
$$\displaystyle \cos^2⁡x={1\over2}+{\cos⁡2x\over2}$$
$$\displaystyle \int \cos^2⁡x \:dx=\int \left({1\over2}+{\cos⁡2x\over2}\right)\: dx$$
$$\displaystyle ={x\over2}+{\sin⁡2x\over4}+c$$

# Odd powers on $$\sin⁡ x$$ and $$\cos ⁡x$$

LESSON 1
Determine
$\int \sin^3⁡x \:dx$
SOLUTION
• We will rewrite $$\int \sin^n⁡x\:dx$$ in the form $$\int \sin ⁡x (\sin^{n-1}⁡x )\:dx$$ where $$n$$ is odd.
$$\int \sin^3⁡x\:dx=\int \sin ⁡x \sin^2⁡x \:dx$$
• Since $$\sin^2⁡x=1-\cos^2⁡x$$
$$\int \sin ⁡x \sin^2⁡x \:dx=\int \sin ⁡x(1-\cos^2⁡x) \:dx$$
$$=\int \sin ⁡x\:dx-\int \sin ⁡x \cos^2⁡x\:dx$$
• We now seek to use $$\displaystyle \int f'(x)[f(x)]^n\:dx={[f(x)]^{n+1}\over n+1}+c$$
$$\int \sin⁡ x \sin^2⁡x\:dx=\int \sin⁡ x \:dx+\int (-\sin ⁡x) \cos^2⁡x\:dx$$
$$f(x)=\cos ⁡x→f'(x)=-\sin ⁡x$$
$$\displaystyle \int \sin ⁡x \sin^2⁡x\:dx=-\cos ⁡x+{\cos^3⁡x\over3}+c$$

LESSON 2
Determine
$\int \cos^3⁡x \:dx$
SOLUTION
• Rewrite $$\int \cos^n⁡x\:dx$$ in the form $$\int \cos ⁡x \cos^2⁡x\: dx$$ where $$n$$ is odd.
$$\int \cos^3⁡x \:dx=\int \cos⁡ x \cos^2⁡x\:dx$$
$$=\int \cos⁡ x (1-\sin^2⁡x)\:dx$$
$$=\int \cos ⁡x\:dx-\int \cos ⁡x \sin^2⁡x \:dx$$
$$\displaystyle =\sin ⁡x-{\sin^3⁡x\over3}+c$$

LESSON 3
Determine
$\int \cos^3⁡x \sin^4⁡x\:dx$
SOLUTION
• Though we have a combination of odd and even powers we will simply manipulate the odd power.
$$\int \cos^3⁡x \sin^4⁡x\:dx=\int \cos ⁡x \cos^2⁡x \sin^4⁡x\:dx$$
$$=\int \cos ⁡x (1-\sin^2⁡x) \sin^4⁡x\:dx$$
$$=\int \cos ⁡x \sin^4⁡x-\int \cos ⁡x \sin^6⁡x \:dx$$
$$\displaystyle ={\sin^5⁡x\over5}-{\sin^7⁡x\over7}+c$$

# Even powers of $$\tan⁡ x$$

LESSON 1
Determine
$\int \tan^2⁡x\:dx$
SOLUTION
$$\int \tan^2⁡x\:dx=\int \sec^2⁡x-1\:dx$$
• Since $$\sec^2⁡x=1+\tan^2⁡x$$
$$\int \tan^2⁡x\:dx=\tan ⁡x-x+c$$