What is the integral of 3^{2x}cos^2(3^x) ?

The integral of 3^{2x}cos^2(3^x) is (2*9^x+2*3^xsin(2*3^x)+cos(2*3^x))/(8ln(3))+C

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integral of 3^{2x}cos^2(3^x)

\int 3^{2 x} cos^{2} (3^{x}) d x

\frac{2 \cdot 9 ^{x} + 2 \cdot 3 ^{x} sin ( 2 \cdot 3 ^{x} ) + cos ( 2 \cdot 3 ^{x} )}{8 ln ( 3 )} + C

Solution steps

\int 3^{2 x} cos^{2} (3^{x}) d x

Apply u-substitution

: \int \frac{u cos ^{2} ( u )}{ln ( 3 )} d u

= \int \frac{u cos ^{2} ( u )}{ln ( 3 )} d u

Take the constant out:

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

= \frac{1}{ln ( 3 )} \cdot \int u cos^{2} (u) d u

Apply Integration By Parts

: \frac{1}{2} u (u + \frac{1}{2} sin (2 u)) - \int \frac{1}{2} (u + \frac{1}{2} sin (2 u)) d u

= \frac{1}{ln ( 3 )} (\frac{1}{2} u (u + \frac{1}{2} sin (2 u)) - \int \frac{1}{2} (u + \frac{1}{2} sin (2 u)) d u)

\int \frac{1}{2} (u + \frac{1}{2} sin (2 u)) d u = \frac{1}{2} (\frac{u ^{2}}{2} - \frac{1}{4} cos (2 u))

= \frac{1}{ln ( 3 )} (\frac{1}{2} u (u + \frac{1}{2} sin (2 u)) - \frac{1}{2} (\frac{u ^{2}}{2} - \frac{1}{4} cos (2 u)))

Substitute back

u = 3^{x}

= \frac{1}{ln ( 3 )} (\frac{1}{2} \cdot 3^{x} (3^{x} + \frac{1}{2} sin (2 \cdot 3^{x})) - \frac{1}{2} (\frac{( 3 ^{x} ) ^{2}}{2} - \frac{1}{4} cos (2 \cdot 3^{x})))

Simplify

\frac{1}{ln ( 3 )} (\frac{1}{2} \cdot 3^{x} (3^{x} + \frac{1}{2} sin (2 \cdot 3^{x})) - \frac{1}{2} (\frac{( 3 ^{x} ) ^{2}}{2} - \frac{1}{4} cos (2 \cdot 3^{x}))) : \frac{2 \cdot 9 ^{x} + 2 \cdot 3 ^{x} sin ( 2 \cdot 3 ^{x} ) + cos ( 2 \cdot 3 ^{x} )}{8 ln ( 3 )}

= \frac{2 \cdot 9 ^{x} + 2 \cdot 3 ^{x} sin ( 2 \cdot 3 ^{x} ) + cos ( 2 \cdot 3 ^{x} )}{8 ln ( 3 )}

Add a constant to the solution

= \frac{2 \cdot 9 ^{x} + 2 \cdot 3 ^{x} sin ( 2 \cdot 3 ^{x} ) + cos ( 2 \cdot 3 ^{x} )}{8 ln ( 3 )} + C

f (x) = x - 3

2 x \frac{d y}{d x} + y = 10 x

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

\frac{d}{d t} (sin (t) cos (t))

\frac{d y}{d x} = 2 y + 2

What is the integral of 3^{2x}cos^2(3^x) ?
The integral of 3^{2x}cos^2(3^x) is (2*9^x+2*3^xsin(2*3^x)+cos(2*3^x))/(8ln(3))+C