Question

Find \(\int sin^2⁡(8x+5) dx\)

(a) \(\frac{x}{4}+\frac{sin⁡(16x+10)}{32}+C\)

(b) \(\frac{x}{2}-\frac{cos⁡(16x+10)}{32}+C\)

(c) \(\frac{x}{2}-\frac{sin⁡(16x+10)}{32}+C\)

(d) \(\frac{x}{2}+\frac{cos⁡(16x+5)}{32}+C\)

The question was asked in unit test.

I'd like to ask this question from Methods of Integration-2 in portion Integrals of Mathematics – Class 12

Answer 1

The correct option is (c) \(\frac{x}{2}-\frac{sin⁡(16x+10)}{32}+C\)

The explanation: \(\int sin^2⁡(8x+5) dx=\int \frac{1-cos⁡2(8x+5)}{2} dx=\int \frac{1}{2} dx-\frac{1}{2} \int cos(16x+10)dx\)

=\(\frac{x}{2}-\frac{1}{2} (\frac{sin⁡(16x+10)}{16})=\frac{x}{2}-\frac{sin⁡(16x+10)}{32}+C\)