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Q12E

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Found in: Page 340

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate the integral$$\int {x*{\mathop{\rm Sin}\nolimits} ({x^2})*{\mathop{\rm Cos}\nolimits} (3{x^2}).dx}$$

The Value of integral is $$- \frac{1}{2}{\mathop{\rm Cos}\nolimits} {x^2} + \frac{3}{2}{{\mathop{\rm Cos}\nolimits} ^2}{x^2} + c$$

See the step by step solution

## Step 1-

Apply the Formula of $${\mathop{\rm Cos}\nolimits} 3x$$

We know,$${\mathop{\rm Cos}\nolimits} 3x = 4{{\mathop{\rm Cos}\nolimits} ^3}x - 3{\mathop{\rm Cos}\nolimits} x$$

Therefore,$$\int {x*{\mathop{\rm Sin}\nolimits} {x^2}} *{\mathop{\rm Cos}\nolimits} 3{x^2}.dx = \int {x*{\mathop{\rm Sin}\nolimits} {x^2}(4{{{\mathop{\rm Cos}\nolimits} }^3}{x^2} - 3{\mathop{\rm Cos}\nolimits} {x^2}).dx}$$

=$$4\int {x*{\mathop{\rm Sin}\nolimits} {x^2}} *{\cos ^3}{x^2} - 3\int {x*{\mathop{\rm Sin}\nolimits} {x^2}*{\mathop{\rm Cos}\nolimits} {x^2}.dx}$$

## Step 2 – Evaluate the integral by substituting $${\mathop{\rm Cos}\nolimits} {x^2} = t$$

Let $${\mathop{\rm Cos}\nolimits} {x^2} = t$$

Differentiating both sides with respect to x

=$$- 2x*{\mathop{\rm Sin}\nolimits} {x^2}.dx = dt$$

=$$x*{\mathop{\rm Sin}\nolimits} {x^2}.dx = ( - )\frac{{dt}}{2}$$

The integral becomes,

=$$4\int {{t^3}( - \frac{{dt}}{2}) - 3\int {t( - \frac{{dt}}{2})} }$$

=$$- 2\int {{t^3}dt + \frac{3}{2}\int {t.dt} }$$

=$$- 2*\frac{{{t^4}}}{4} + \frac{3}{2}*\frac{{{t^2}}}{2} + c$$

=$$- \frac{{{t^4}}}{2} + \frac{3}{4}*\frac{{{t^2}}}{{}} + c$$

## Step 3 –(Put $$t = {\mathop{\rm Cos}\nolimits} {x^2}$$in the value of integral)=$$- \frac{1}{2}{{\mathop{\rm Cos}\nolimits} ^4}{x^2} + \frac{3}{2}{{\mathop{\rm Cos}\nolimits} ^2}{x^2} + c$$

Hence, $$\int {x*{\mathop{\rm Sin}\nolimits} {x^2}} *{\mathop{\rm Cos}\nolimits} 3{x^2}.dx$$= $$- \frac{1}{2}{{\mathop{\rm Cos}\nolimits} ^4}{x^2} + \frac{3}{2}{{\mathop{\rm Cos}\nolimits} ^2}{x^2} + c$$