What is the integral of 7sin^6(x)cos^3(x) ?

The integral of 7sin^6(x)cos^3(x) is sin^7(x)-7/9 sin^9(x)+C

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integral of 7sin^6(x)cos^3(x)

\int 7 sin^{6} (x) cos^{3} (x) d x

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

Solution steps

\int 7 sin^{6} (x) cos^{3} (x) d x

Take the constant out:

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

= 7 \cdot \int sin^{6} (x) cos^{3} (x) d x

Rewrite using trig identities

= 7 \cdot \int (1 - sin^{2} (x)) cos (x) sin^{6} (x) d x

Apply u-substitution

: \int u^{6} (1 - u^{2}) d u

= 7 \cdot \int u^{6} (1 - u^{2}) d u

Expand

u^{6} (1 - u^{2}) : u^{6} - u^{8}

= 7 \cdot \int u^{6} - u^{8} 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

= 7 (\int u^{6} d u - \int u^{8} d u)

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

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

= 7 (\frac{u ^{7}}{7} - \frac{u ^{9}}{9})

Substitute back

u = sin (x)

= 7 (\frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9})

Simplify

7 (\frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9}) : sin^{7} (x) - \frac{7}{9} sin^{9} (x)

= sin^{7} (x) - \frac{7}{9} sin^{9} (x)

Add a constant to the solution

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

x \to 5 lim (7)

(2 x - y) d x + (2 y - x) d y = 0, y (1) = 3

csc (x)

\int \frac{1 - sin ( θ )}{θ + cos ( θ )} d θ

\int e^{- x} - sin (2 x) d x

What is the integral of 7sin^6(x)cos^3(x) ?
The integral of 7sin^6(x)cos^3(x) is sin^7(x)-7/9 sin^9(x)+C