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

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

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integral of sin^4(x)cos^3(x)

\int sin^{4} (x) cos^{3} (x) d x

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

Solution steps

\int sin^{4} (x) cos^{3} (x) d x

Rewrite using trig identities

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

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 = sin (x)

= \frac{sin ^{5} ( x )}{5} - \frac{sin ^{7} ( x )}{7}

Add a constant to the solution

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

\int \frac{y ^{2} + 1}{( y - 1 ) ^{2} ( y + 1 )} d y

y^{^{''}} + 2 y^{^{'}} + 4 y = e^{t} cos (t)

\frac{\partial}{\partial z} (y sin (x z))

t \to - 1 lim (\frac{t ^{10} + t ^{5} + 1}{t - 1})

1936 - 16 t^{2}

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