What is the integral of (1+sin(t))^9cos(t) ?

The integral of (1+sin(t))^9cos(t) is ((1+sin(t))^{10})/(10)+C

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integral of (1+sin(t))^9cos(t)

\int (1 + sin (t))^{9} cos (t) d t

\frac{( 1 + sin ( t ) ) ^{10}}{10} + C

Solution steps

\int (1 + sin (t))^{9} cos (t) d t

Apply u-substitution

= \int u^{9} d u

Apply the Power Rule

= \frac{u ^{10}}{10}

Substitute back

u = 1 + sin (t)

= \frac{( 1 + sin ( t ) ) ^{10}}{10}

Add a constant to the solution

= \frac{( 1 + sin ( t ) ) ^{10}}{10} + C

What is the integral of (1+sin(t))^9cos(t) ?
The integral of (1+sin(t))^9cos(t) is ((1+sin(t))^{10})/(10)+C