What is the integral of 42cos^2(97x) ?

The integral of 42cos^2(97x) is 21(x+1/194 sin(194x))+C

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integral of 42cos^2(97x)

\int 42 cos^{2} (97 x) d x

21 (x + \frac{1}{194} sin (194 x)) + C

Solution steps

\int 42 cos^{2} (97 x) d x

Take the constant out:

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

= 42 \cdot \int cos^{2} (97 x) d x

Rewrite using trig identities

= 42 \cdot \int \frac{1}{2} (1 + cos (194 x)) d x

Take the constant out:

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

= 42 \cdot \frac{1}{2} \cdot \int 1 + cos (194 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 42 \cdot \frac{1}{2} (\int 1 d x + \int cos (194 x) d x)

\int 1 d x = x

\int cos (194 x) d x = \frac{1}{194} sin (194 x)

= 42 \cdot \frac{1}{2} (x + \frac{1}{194} sin (194 x))

Simplify

42 \cdot \frac{1}{2} (x + \frac{1}{194} sin (194 x)) : 21 (x + \frac{1}{194} sin (194 x))

= 21 (x + \frac{1}{194} sin (194 x))

Add a constant to the solution

= 21 (x + \frac{1}{194} sin (194 x)) + C

y^{^{''}} - 0.2 y^{^{'}} + 0.01 y = 0, y (0) = 15, y^{^{'}} (0) = b

x d y - (2 x + 1) e^{- y} d x = 0

\int \frac{1}{( 3 x - 1 )} d x

\frac{\partial}{\partial x} (2 cos (3 x))

\frac{d y}{d x} + y^{5} x + 4 y = 0

What is the integral of 42cos^2(97x) ?
The integral of 42cos^2(97x) is 21(x+1/194 sin(194x))+C