What is the integral of tcos^2(2t) ?

The integral of tcos^2(2t) is 1/32 (8t^2+4tsin(4t)+cos(4t))+C

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integral of tcos^2(2t)

\int t cos^{2} (2 t) d t

\frac{1}{32} (8 t^{2} + 4 t sin (4 t) + cos (4 t)) + C

Solution steps

\int t cos^{2} (2 t) d t

Rewrite using trig identities

= \int \frac{1}{2} t (1 + cos (4 t)) d t

Take the constant out:

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

= \frac{1}{2} \cdot \int t (1 + cos (4 t)) d t

Apply Integration By Parts

: t (t + \frac{1}{4} sin (4 t)) - \int t + \frac{1}{4} sin (4 t) d t

= \frac{1}{2} (t (t + \frac{1}{4} sin (4 t)) - \int t + \frac{1}{4} sin (4 t) d t)

\int t + \frac{1}{4} sin (4 t) d t = \frac{t ^{2}}{2} - \frac{1}{16} cos (4 t)

= \frac{1}{2} (t (t + \frac{1}{4} sin (4 t)) - (\frac{t ^{2}}{2} - \frac{1}{16} cos (4 t)))

Simplify

\frac{1}{2} (t (t + \frac{1}{4} sin (4 t)) - (\frac{t ^{2}}{2} - \frac{1}{16} cos (4 t))) : \frac{1}{32} (8 t^{2} + 4 t sin (4 t) + cos (4 t))

= \frac{1}{32} (8 t^{2} + 4 t sin (4 t) + cos (4 t))

Add a constant to the solution

= \frac{1}{32} (8 t^{2} + 4 t sin (4 t) + cos (4 t)) + C

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What is the integral of tcos^2(2t) ?
The integral of tcos^2(2t) is 1/32 (8t^2+4tsin(4t)+cos(4t))+C