Math  /  Calculus

Questionsin6xcos3xdx\int \sin ^{6} x \cdot \cos ^{3} x d x

Studdy Solution

STEP 1

1. The integral involves powers of sine and cosine functions.
2. We will use trigonometric identities and substitution to simplify the integral.

STEP 2

1. Simplify the integral using trigonometric identities.
2. Use substitution to solve the integral.
3. Integrate and back-substitute.

STEP 3

First, we observe that the integral involves sin6x\sin^6 x and cos3x\cos^3 x. We can use the identity sin2x=1cos2x\sin^2 x = 1 - \cos^2 x to simplify sin6x\sin^6 x:
sin6x=(sin2x)3=(1cos2x)3\sin^6 x = (\sin^2 x)^3 = (1 - \cos^2 x)^3
Rewrite the integral:
sin6xcos3xdx=(1cos2x)3cos3xdx\int \sin^6 x \cdot \cos^3 x \, dx = \int (1 - \cos^2 x)^3 \cdot \cos^3 x \, dx

STEP 4

Now, use substitution. Let u=cosx u = \cos x , then du=sinxdx du = -\sin x \, dx , or sinxdx=du \sin x \, dx = -du .
Rewrite the integral in terms of u u :
(1u2)3u3(du)\int (1 - u^2)^3 \cdot u^3 \cdot (-du)
This simplifies to:
(1u2)3u3du-\int (1 - u^2)^3 \cdot u^3 \, du

STEP 5

Expand (1u2)3(1 - u^2)^3 using the binomial theorem:
(1u2)3=13u2+3u4u6(1 - u^2)^3 = 1 - 3u^2 + 3u^4 - u^6
Substitute back into the integral:
(13u2+3u4u6)u3du-\int (1 - 3u^2 + 3u^4 - u^6) \cdot u^3 \, du
Distribute u3u^3:
(u33u5+3u7u9)du-\int (u^3 - 3u^5 + 3u^7 - u^9) \, du
Integrate each term separately:
=(u443u66+3u88u1010)+C= -\left( \frac{u^4}{4} - \frac{3u^6}{6} + \frac{3u^8}{8} - \frac{u^{10}}{10} \right) + C
Simplify:
=(u44u62+3u88u1010)+C= -\left( \frac{u^4}{4} - \frac{u^6}{2} + \frac{3u^8}{8} - \frac{u^{10}}{10} \right) + C

STEP 6

Back-substitute u=cosx u = \cos x :
=(cos4x4cos6x2+3cos8x8cos10x10)+C= -\left( \frac{\cos^4 x}{4} - \frac{\cos^6 x}{2} + \frac{3\cos^8 x}{8} - \frac{\cos^{10} x}{10} \right) + C
The solution to the integral is:
(cos4x4cos6x2+3cos8x8cos10x10)+C-\left( \frac{\cos^4 x}{4} - \frac{\cos^6 x}{2} + \frac{3\cos^8 x}{8} - \frac{\cos^{10} x}{10} \right) + C

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