What is the integral of sin^3(θ)cos^4(θ) ?

The integral of sin^3(θ)cos^4(θ) is -(cos^5(θ))/5+(cos^7(θ))/7+C

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integral of sin^3(θ)cos^4(θ)

\int sin^{3} (θ) cos^{4} (θ) d θ

- \frac{cos ^{5} ( θ )}{5} + \frac{cos ^{7} ( θ )}{7} + C

Solution steps

\int sin^{3} (θ) cos^{4} (θ) d θ

Rewrite using trig identities

= \int (1 - cos^{2} (θ)) sin (θ) cos^{4} (θ) d θ

Apply u-substitution

: \int - u^{4} (1 - u^{2}) d u

= \int - u^{4} (1 - u^{2}) d u

Expand

- u^{4} (1 - u^{2}) : - u^{4} + u^{6}

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

\int u^{4} d u = \frac{u ^{5}}{5}

\int u^{6} d u = \frac{u ^{7}}{7}

= - \frac{u ^{5}}{5} + \frac{u ^{7}}{7}

Substitute back

u = cos (θ)

= - \frac{cos ^{5} ( θ )}{5} + \frac{cos ^{7} ( θ )}{7}

Add a constant to the solution

= - \frac{cos ^{5} ( θ )}{5} + \frac{cos ^{7} ( θ )}{7} + C

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

\int (5 e^{x}) d x

f (x) = \frac{1}{x - 2}, at x = 7

\int \frac{x}{36 x ^{2} - 1} d x

f (x) = 6 e^{x} + \frac{2}{3 x}

What is the integral of sin^3(θ)cos^4(θ) ?
The integral of sin^3(θ)cos^4(θ) is -(cos^5(θ))/5+(cos^7(θ))/7+C