Question

Q3. The integral value of S sin 2x sin x dx is​

Answer 1

Step-by-step explanation:

∫

0

2

π

sin2xsinxdx=K∫

0

2

π

cos2xcosxdx

∫

0

2

π

2sinxcosxsinxdx=K∫

0

2

π

(1−2sin

2

x)cosxdx

∫

0

2

π

2sin

2

xcosxdx=K

⎣

⎢

⎢

⎡

∫

0

2

π

cosxdx−∫

0

2

π

2sin

2

xcosxdx

⎦

⎥

⎥

⎤

{

3

2sin

3

x

}

0

2

π

=K

⎣

⎢

⎢

⎡

{sinx}

0

2

π

−{

3

2sin

3

x

}

0

2

π

⎦

⎥

⎥

⎤

3

2

=K[1−

3

2

]

Answer 2

SOLUTION

TO EVALUATE

$\displaystyle \sf{\int\limits_{}^{} \sin 2x \sin x \: \, dx}$

EVALUATION

We have to find the value of

$\displaystyle \sf{\int\limits_{}^{} \sin 2x \sin x \: \, dx}$

Here

$\displaystyle \sf{ \sin 2x \sin x \: }$

$\displaystyle \sf{ = \frac{1}{2} \times 2 \sin 2x \sin x \: }$

$\displaystyle \sf{ = \frac{1}{2} \bigg[\cos (2x - x) - \cos (2x + x) \bigg]}$

$\displaystyle \sf{ = \frac{1}{2} \bigg[\cos x - \cos 3x \bigg]}$

Now

$\displaystyle \sf{\int\limits_{}^{} \sin 2x \sin x \: \, dx}$

$\displaystyle \sf{ = \int\limits_{}^{} \frac{1}{2} \bigg[\cos x - \cos 3x \bigg] \: \, dx}$

$\displaystyle \sf{ = \frac{1}{2} \int\limits_{}^{} \bigg[\cos x - \cos 3x \bigg] \: \, dx}$

$\displaystyle \sf{ = \frac{1}{2} \int\limits_{}^{} \cos x \, dx} - \displaystyle \sf{ \frac{1}{2} \int\limits_{}^{} \cos 3x \: \, dx}$

$\displaystyle \sf{ = \frac{1}{2} \sin x - \frac{1}{6} \sin 3x + c }$

Where c is integration constant

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