What is the integral of cos(x)(6+4sin^2(x)) ?

The integral of cos(x)(6+4sin^2(x)) is 6sin(x)+(4sin^3(x))/3+C

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integral of cos(x)(6+4sin^2(x))

\int cos (x) (6 + 4 sin^{2} (x)) d x

6 sin (x) + \frac{4 sin ^{3} ( x )}{3} + C

Solution steps

\int cos (x) (6 + 4 sin^{2} (x)) d x

Apply u-substitution

: \int 6 + 4 u^{2} d u

= \int 6 + 4 u^{2} d u

Apply the Sum Rule:

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

= \int 6 d u + \int 4 u^{2} d u

\int 6 d u = 6 u

\int 4 u^{2} d u = \frac{4 u ^{3}}{3}

= 6 u + \frac{4 u ^{3}}{3}

Substitute back

u = sin (x)

= 6 sin (x) + \frac{4 sin ^{3} ( x )}{3}

Add a constant to the solution

= 6 sin (x) + \frac{4 sin ^{3} ( x )}{3} + C

\frac{\partial}{\partial x} (9 - 3 x^{2} - y^{2})

x \to 2 lim (3 - x^{2})

n = 0 \sum \infty 3

y^{^{'}} = y sin (2 x)

\int (5 x - 1) (3 x^{3} + 2) d x

What is the integral of cos(x)(6+4sin^2(x)) ?
The integral of cos(x)(6+4sin^2(x)) is 6sin(x)+(4sin^3(x))/3+C