Evaluate the indefinite integral. ∫cos⁡x(3sin⁡x−1)dx

Question

Evaluate the indefinite integral.  ∫cos⁡x(3sin⁡x−1)dx

scomparve5j · Accepted Answer

Step 1  Consider the provided integral,  ∫cos⁡x(3sin⁡x−1)dx  Evaluate the indefinite integral.  Apply the substitution method,  Let, u=3sin⁡x−1⇒du=3cos⁡xdx.  Step 2  Therefore the integral becomes,  ∫cos⁡x(3sin⁡x−1)dx=∫u3du  =13⋅∫udu  =13⋅u1+11+1+C  =u26+C  Step 3  Substitute back u=3sin⁡x−1 in the above integral,  ∫cos⁡x(3sin⁡x−1)dx=16(3sin⁡(x)−1)2+C  Hence.

Cheryl King · Accepted Answer

∫cos⁡(x)(3sin⁡(x)−1)dx  =13∫udu  ∫udu  =u22  13∫udu  =u26  =(3sin⁡(x)−1)26  ∫cos⁡(x)(3sin⁡(x)−1)dx  =(3sin⁡(x)−1)26+C

karton · Accepted Answer

∫cos⁡(x)×(3sin⁡(x)−1)dx Remove the parentheses ∫3cos⁡(x)sin⁡(x)−cos⁡(x)dx Simplify the expression ∫32×sin⁡(2x)−cos⁡(x)dx Calculate ∫3sin⁡(2x)2−cos⁡(x)dx Use properties of integrals ∫3sin⁡(2x)2dx−∫cos⁡(x)dx Evaluate the integrals −3cos⁡(2x)4−sin⁡(x) Add C Solution −3cos⁡(2x)4−sin⁡(x)+C

Evaluate the indefinite integral. \int \cos x(3\sin x-1)dx

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