Evaluate ∫ cos33x dx

1.	Evaluate ∫ cos33x dx
Answer» We can write ∫ cos³3x dx as: ∫cos3x(cos3x)²dx ∫cos³x(cos²3x)dx and further as : = cos3x(1-sin²3x)dx = ∫cos3xdx-∫cos3x(sin²3x)dx Taking A = ∫cos3xdx Solving for A, A = \(\frac{sin3x}{3}\) Taking B = ∫cos3x(sin²3x)dx In this taking sin3x = t Differentiating on both sides we get, 3cos³xdx = dt Solving by putting these values we get, B = \(\int\frac{t^2}{3}\)dt = \(\frac{t^3}{9}\) + c Substituting values we get, B = \(\frac{sin^33x}{9}\) + c Our final answer is A+B i.e, \(\frac{sin3x}{3}\) + \(\frac{sin3x}{3}\) + c

Answer»

We can write ∫ cos³3x dx as:

∫cos3x(cos3x)²dx ∫cos³x(cos²3x)dx and

further as :

= cos3x(1-sin²3x)dx

= ∫cos3xdx-∫cos3x(sin²3x)dx

Taking A = ∫cos3xdx

Solving for A,

A = \(\frac{sin3x}{3}\)

Taking B = ∫cos3x(sin²3x)dx

In this taking sin3x = t

Differentiating on both sides we get,

3cos³xdx = dt

Solving by putting these values we get,

B = \(\int\frac{t^2}{3}\)dt

= \(\frac{t^3}{9}\) + c

Substituting values we get,

B = \(\frac{sin^33x}{9}\) + c

Our final answer is A+B i.e,

\(\frac{sin3x}{3}\) + \(\frac{sin3x}{3}\) + c

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