Use the Table of Integrals toevaluate the integral ∫cos⁡xsin2x−9dx

Question

Use the Table of Integrals  toevaluate the integral  ∫cos⁡xsin2x−9dx

enhebrevz · Accepted Answer

Step 1
Consider the following integral:

\int \frac{\cos (x)}{\sin^{2} (x) - 9} d x

Substitute

\sin (x) = u \Rightarrow \cos (x) d x = d u

in the above integral:

\int \frac{\cos (x)}{\sin^{2} (x) - 9} d x = \int \frac{1}{u^{2} - 9} d u

= \int \frac{1}{(u + 3) (u - 3)} d u

= \frac{1}{6} \int (\frac{1}{u - 3} - \frac{1}{u + 3}) d u

Step 2
Use the

\int \frac{1}{x} d x = \ln | x | + C

formula:

\int \frac{\cos (x)}{\sin^{2} (x) - 9} d x = \frac{1}{6} \ln | u - 3 | - \frac{1}{6} \ln | u + 3 | + C

Substitute

u = \sin (x)

in the above equation:

\int \frac{\cos (x)}{\sin^{2} (x) - 9} d x = \frac{1}{6} \ln | \sin (x) - 3 | - \frac{1}{6} \ln | \sin (x) + 3 | + C

= \frac{1}{6} \ln [3 - \sin (x)] - \frac{1}{6} \ln [\sin (x) + 3] + C

Step 3
Hence, the solution is

\frac{1}{6} \ln [3 - \sin (x)] - \frac{1}{6} \ln [\sin (x) + 3] + C

.

Dabanka4v · Accepted Answer

Use the Table of Integrals toevaluate the integral \int \frac{\cos