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

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

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integral of 7sin^2(x)

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

\frac{7}{2} (x - \frac{1}{2} sin (2 x)) + C

Solution steps

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

Take the constant out:

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

= 7 \cdot \int sin^{2} (x) d x

Rewrite using trig identities

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

Take the constant out:

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

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

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

\int 1 d x = x

\int cos (2 x) d x = \frac{1}{2} sin (2 x)

= 7 \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x))

Simplify

7 \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x)) : \frac{7}{2} (x - \frac{1}{2} sin (2 x))

= \frac{7}{2} (x - \frac{1}{2} sin (2 x))

Add a constant to the solution

= \frac{7}{2} (x - \frac{1}{2} sin (2 x)) + C

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\int \frac{6 x}{( x - 5 ) ^{2}} d x

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\int_{- \infty}^{0} e^{9 x} d x

t^{2} (\frac{d y}{d t}) + 4 t y - y^{3} = 0

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