Evaluate the integrals in Exercises 1–22.
∫ 7cos⁷(t) dt

Name: Evaluate the integral: ∫5{{{\displaystyle\int5}}} sin7x\sin^7x dxdx
Uploaded: 2026-01-15T08:21:58.091Z
Duration: 5 min
Description: Watch the video explanation for the concept: Evaluate the integral: ∫5{{{\displaystyle\int5}}} sin7x\sin^7x dxdx

Question

Evaluate the integrals in Exercises 1–22.

∫ 7cos⁷(t) dt

Accepted Answer

Welcome back everyone. Evaluate the integral 5 signs of the power of 7 of x d x. For this problem, let's begin by factoring out the constants, so we get 5 integral of sin to the power of 7 of x d x, and we basically want to integrate sin to the power of 7 of x with respect to x. What we're going to do is basically rewrite signs to the power of 7 using the properties of the exponents. So we're going to get 5 integral of. First of all, we can write sin to the power of 6 of x multiplied by sinx followed by dx, right? And now this is going to be equal to 5 integral of sine. Cubed of x squared or alternatively we can write sin squared of x cubed using the properties of exponents multiplied by sin x dx. Now let's remember the Pythagorean identity. What we can do is rewrite sin squared that's 1 minus cosine squared, so that'll be 5 integral of 1 minus cosine squared of x cubed multiplied by sin x dx. And now why is this useful? Well, what we can do is simply introduce substitution. Let's suppose that u is equal to cosine x. DU divided by DX is going to be the derivative of cosine x. Which is negative sin x. Multiplying both sides by negative dx, we get negative duu is equal to sin x dx, which is exactly what we have. Sinx dx within the integrant will be equal to negative duu, and we can now rewrite the integral as 5, integral of 1 minus u square cubed, multiplied by duu. So we can just write DU and factor out the negative sign. Well done. So that's -5 integral of. We're going to Cube 1 minus u 2 and this is equal to 1 minus 3 u2 plus 3 u 4 minus u of 6. Well then from here we can integrate, right? Essentially, to integrate, we're going to apply the power rule for each term and we have -5 in front. So we get -5 in parentheses. The integral of 1 duu is simply the integral of duu, which is u minus. 3, the integral of u squared is u cubed divided by 3 + 3. The integral of u to the power of 4 is u to the power of 5 divided by 5 minus the integral of u to the power of 6 is c u to the power of 7 divided by 7, and we're going to add a constant of integration c. And from here we're going to distribute and replace you with cosine, because u is cosine x. So we get -5U which is -5 cosine x, that's our first term. -5 multiplied by -3 divided by 3 is -5 multiplied by -1, that's 1. So our next term is going to be a cube. Of you multiplied by 5, we're going to add 5. Cosine cubed of x. Then we have -5 divided by 5, which is -1 multiplied by 3, that's -3. So we're going to subtract 3 u to the power of 5 is cosine to the power of 5 of x. And finally we get -5 divided by -7, which is plus 5 divided by 7 ys the power of 7. So that would be cosine to the power of 7 of x plus a constant of integration c. And that's our final answer for this problem. Thank you for watching.

Evaluate the integrals in Exercises 1–22.∫ 7cos⁷(t) dt

Watch next

Evaluate the integrals in Exercises 1–22.
∫ 7cos⁷(t) dt