What is the integral of 4sin^3(x)cos^7(x) ?

The integral of 4sin^3(x)cos^7(x) is 4(-(cos^8(x))/8+(cos^{10}(x))/(10))+C

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integral of 4sin^3(x)cos^7(x)

\int 4 sin^{3} (x) cos^{7} (x) d x

4 (- \frac{cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10}) + C

Solution steps

\int 4 sin^{3} (x) cos^{7} (x) d x

Take the constant out:

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

= 4 \cdot \int sin^{3} (x) cos^{7} (x) d x

Rewrite using trig identities

= 4 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{7} (x) d x

Apply u-substitution

: \int - u^{7} (1 - u^{2}) d u

= 4 \cdot \int - u^{7} (1 - u^{2}) d u

Expand

- u^{7} (1 - u^{2}) : - u^{7} + u^{9}

= 4 \cdot \int - u^{7} + u^{9} 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

= 4 (- \int u^{7} d u + \int u^{9} d u)

\int u^{7} d u = \frac{u ^{8}}{8}

\int u^{9} d u = \frac{u ^{10}}{10}

= 4 (- \frac{u ^{8}}{8} + \frac{u ^{10}}{10})

Substitute back

u = cos (x)

= 4 (- \frac{cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10})

Add a constant to the solution

= 4 (- \frac{cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10}) + C

\int (x^{2} - y^{2}) d x

x \to 1 lim (ln (5 x) + e^{x})

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

y^{^{''}} - 36 y = 0

\frac{d}{d t} (\frac{2 t}{1 + t ^{2}})

What is the integral of 4sin^3(x)cos^7(x) ?
The integral of 4sin^3(x)cos^7(x) is 4(-(cos^8(x))/8+(cos^{10}(x))/(10))+C