Get the answer to "1.∫{{sin⁡23xdx 2.∫{{xcos⁡6x2dx 3.∫{{tan⁡9xdx 4,∫{{x2cot⁡x3dx"

Monincbh

Answered question

2021-12-03

1 . \int_{{}^{{} \sin \frac{2}{3} x d x

2 . \int_{{}^{{} x \cos 6 x^{2} d x

3 . \int_{{}^{{} \tan 9 x d x

4, \int_{{}^{{} x^{2} {\cot x}^{3} d x

Answer & Explanation

Steven Arredondo

Beginner2021-12-04Added 18 answers

Solution of question 1
$\int_{{}^{{} \sin (\frac{2}{3} x) d x$
Put $u = \frac{2 x}{3} \Rightarrow \frac{d u}{d x} = \frac{2}{3} or \frac{3 d u}{2} = d x$
$\Rightarrow \int_{{}^{{} \sin u \frac{3}{2} d u = \frac{3}{2} \int \sin u d u$
Now solving integral
$= \frac{3}{2} - \cos (u) + c$
$= \frac{- 3 \cos u}{2} + c$
$= \frac{- 3 \cos (\frac{2}{3 x})}{2} + c$
Put back $u = (\frac{2}{3}) x$
Hence above is the required solution pf given integral
Solution of question 3
$\int_{{}^{{} \times \cos (6 x^{2}) d x$
Put $u = 6 x^{2} \Rightarrow \frac{d u}{d x} = 12 x \Rightarrow d x = \frac{1}{12 x} d u$
$\Rightarrow \int_{{}^{{} x \cos u \frac{1}{12 x} d u$
$\Rightarrow \frac{1}{12} \int_{{}^{{} \cos u d u$
Now solving integral
$\Rightarrow \frac{1}{12} \sin (u) + c$
Put back $u = 6 x^{2}$
$\Rightarrow \frac{\sin 6 x^{2}}{12} + c$
Hence above is the required solution of given integral
Solution of question 3
$\int_{{}^{{} \tan 9 x$
Put $u = 9 x \Rightarrow \frac{d u}{d x} = 0 a \Rightarrow u = \frac{1}{9} d u$
$\Rightarrow \frac{1}{0} \int_{{}^{{} \tan (u) d x$
Now solving integral
Rewosite $\tan (u) = \frac{\sin u}{\cos u}$
$\Rightarrow \int_{{}^{{} \frac{1}{9}$

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