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

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

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

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

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

Solution steps

\int 3 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

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

Rewrite using trig identities

= 3 \cdot \int (1 - cos^{2} (x)) sin (x) cos^{7} (x) d x

Apply u-substitution

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

= 3 \cdot \int - u^{7} (1 - u^{2}) d u

Expand

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

= 3 \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

= 3 (- \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}

= 3 (- \frac{u ^{8}}{8} + \frac{u ^{10}}{10})

Substitute back

u = cos (x)

= 3 (- \frac{cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10})

Add a constant to the solution

= 3 (- \frac{cos ^{8} ( x )}{8} + \frac{cos ^{10} ( x )}{10}) + C

x = 16 - y^{2}, x = y^{2} - 16

n = 6 \sum \infty \frac{3 n}{12 n}

y^{^{'''}} + y^{^{'}} = 2 x^{2}

f (x) = e^{t a n (x)}

\frac{\partial}{\partial u} (\frac{v}{1 + uv w})

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