What is the integral of sin^4(x)cos^7(x) ?

The integral of sin^4(x)cos^7(x) is (sin^5(x))/5-(3sin^7(x))/7+(sin^9(x))/3-(sin^{11}(x))/(11)+C

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integral of sin^4(x)cos^7(x)

\int sin^{4} (x) cos^{7} (x) d x

\frac{sin ^{5} ( x )}{5} - \frac{3 sin ^{7} ( x )}{7} + \frac{sin ^{9} ( x )}{3} - \frac{sin ^{11} ( x )}{11} + C

Solution steps

\int sin^{4} (x) cos^{7} (x) d x

Rewrite using trig identities

= \int (1 - sin^{2} (x))^{3} cos (x) sin^{4} (x) d x

Apply u-substitution

: \int u^{4} (1 - u^{2})^{3} d u

= \int u^{4} (1 - u^{2})^{3} d u

Expand

u^{4} (1 - u^{2})^{3} : u^{4} - 3 u^{6} + 3 u^{8} - u^{10}

= \int u^{4} - 3 u^{6} + 3 u^{8} - u^{10} 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 3 u^{6} d u + \int 3 u^{8} d u - \int u^{10} d u

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

\int 3 u^{6} d u = \frac{3 u ^{7}}{7}

\int 3 u^{8} d u = \frac{u ^{9}}{3}

\int u^{10} d u = \frac{u ^{11}}{11}

= \frac{u ^{5}}{5} - \frac{3 u ^{7}}{7} + \frac{u ^{9}}{3} - \frac{u ^{11}}{11}

Substitute back

u = sin (x)

= \frac{sin ^{5} ( x )}{5} - \frac{3 sin ^{7} ( x )}{7} + \frac{sin ^{9} ( x )}{3} - \frac{sin ^{11} ( x )}{11}

Add a constant to the solution

= \frac{sin ^{5} ( x )}{5} - \frac{3 sin ^{7} ( x )}{7} + \frac{sin ^{9} ( x )}{3} - \frac{sin ^{11} ( x )}{11} + C

\int 4 x - 11 d x

y^{^{'}} = e^{4 c o s (x)}

2

x \to 6 - lim (36 - x^{2})

\int_{2}^{\infty} 1 d x

What is the integral of sin^4(x)cos^7(x) ?
The integral of sin^4(x)cos^7(x) is (sin^5(x))/5-(3sin^7(x))/7+(sin^9(x))/3-(sin^{11}(x))/(11)+C