What is the integral of sin^2(6x+7) ?

The integral of sin^2(6x+7) is -xcos^2(6x+7)-1/24 sin(2(6x+7))+1/2 xcos(2(6x+7))+x+C

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integral of sin^2(6x+7)

\int sin^{2} (6 x + 7) d x

- x cos^{2} (6 x + 7) - \frac{1}{24} sin (2 (6 x + 7)) + \frac{1}{2} x cos (2 (6 x + 7)) + x + C

Solution steps

\int sin^{2} (6 x + 7) d x

Rewrite using trig identities

= \int \frac{1 - cos ( 2 ( 6 x + 7 ) )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int 1 - cos (2 (6 x + 7)) d x

Rewrite using trig identities

= \frac{1}{2} \cdot \int - 2 cos^{2} (6 x + 7) + 2 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int 2 cos^{2} (6 x + 7) d x + \int 2 d x)

\int 2 cos^{2} (6 x + 7) d x = 2 (x cos^{2} (6 x + 7) + \frac{1}{24} (- 12 x cos (2 (6 x + 7)) + sin (2 (6 x + 7))))

\int 2 d x = 2 x

= \frac{1}{2} (- 2 (x cos^{2} (6 x + 7) + \frac{1}{24} (- 12 x cos (2 (6 x + 7)) + sin (2 (6 x + 7)))) + 2 x)

Simplify

\frac{1}{2} (- 2 (x cos^{2} (6 x + 7) + \frac{1}{24} (- 12 x cos (2 (6 x + 7)) + sin (2 (6 x + 7)))) + 2 x) : - x cos^{2} (6 x + 7) - \frac{1}{24} sin (2 (6 x + 7)) + \frac{1}{2} x cos (2 (6 x + 7)) + x

= - x cos^{2} (6 x + 7) - \frac{1}{24} sin (2 (6 x + 7)) + \frac{1}{2} x cos (2 (6 x + 7)) + x

Add a constant to the solution

= - x cos^{2} (6 x + 7) - \frac{1}{24} sin (2 (6 x + 7)) + \frac{1}{2} x cos (2 (6 x + 7)) + x + C

\int - 18 e^{- 0.6 t} d t

\int x^{3} (x^{4} + 16)^{6} d x

\int \frac{1}{2 x ^{2} - 6 x + 5} d x

\int t sec^{2} (t^{2}) tan^{4} (t^{2}) d t

\int \frac{1}{6 x ^{2} + 3 x + 1} d x

What is the integral of sin^2(6x+7) ?
The integral of sin^2(6x+7) is -xcos^2(6x+7)-1/24 sin(2(6x+7))+1/2 xcos(2(6x+7))+x+C