How do you find the integral of #int cos^5xsin^4xdx#?

Answer 1

Refer to explanation

We possess that

#int cos^5xsin^4xdx=intcosxcos^4xsin^4xdx= intcosx(1-sin^2x)*(1-sin^2x)*sin^4xdx#

We set #sinx=t# hence #dxcosx=dt# so we have

#int(1-t^2)(1-t^2)*t^4dt=int(t^4+t^8-2t^6)dt= t^5/5+t^9/9-2y^7/7+c#

Hence we replace with #sinx# to get
#int cos^5xsin^4xdx=sin^5x/5+sin^9x/9-2*sin^7x/7+c#
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Answer 2

To find the integral of (\int \cos^5(x) \sin^4(x) , dx), you can use trigonometric identities to simplify the integral. Start by using the power-reducing identities:

[ \cos^2(x) = \frac{1 + \cos(2x)}{2} \quad \text{and} \quad \sin^2(x) = \frac{1 - \cos(2x)}{2} ]

Then use the identity (\sin(2x) = 2 \sin(x) \cos(x)) to further simplify. After that, you may use a substitution to make the integral more manageable. Once simplified, integrate the expression and then apply the reverse of your substitution to find the final result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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