Question

ʜʟᴏ ǫᴜᴇ...ɪɴᴛᴇɢʀᴀᴛᴇ....




ɴᴏ ɪʀʀᴇʟᴀᴠᴀɴᴛ ᴀɴs ɴᴇᴇᴅᴇᴅ .....✌️✌️



"ǫᴜᴇ ᴀsᴋᴇᴅ ʙʏ sᴀᴍ"

Answer 1

Answer

• √3/4 - π/12

Given

$\bullet \;\;\bf \int\limits^{\frac{\sqrt{3}}{4}}_0 {\dfrac{2x.sin^{-1}2x}{\sqrt{1-4x^2}}} \, dx$

To Find

• Solve the integral

Solution

$\tt \int {\dfrac{2x.sin^{-1}2x}{\sqrt{1-4x^2}}} \, dx\\\\\tt Let,t=sin^{-1}2x\;\; \& \;\; sint=2x\\\\\implies dt=\dfrac{2}{\sqrt{1-4x^2}}dx\;\; \& \;\; x=\dfrac{sint}{2}$

$\tt \int \dfrac{sint}{2}tdt\\\\\implies \sf \dfrac{1}{2}\int t.sintdt ...(1)[\; Apply\ ILATE\ rule \;]$

Then ,

$\implies \tt u=t\;\;\&\;\;dv=sintdt\\\\\implies \tt du=dt \;\; \& \;\; v=-cost$

A/c to ILATE rule ,

uv - ∫ vdu

⇒ - t cos t - ∫ - cos t dt

⇒ - t cos t + ∫ cos t dt

⇒ - t cos t + sin t

⇒ sin t - t cos t

Now sub. all of above values .

$\implies \tt \dfrac{1}{2}\big[2x-sin^{-1}x\sqrt{1-4x^2}\;\big](From\ (1))$

So , finally we get ,

$\bullet \;\; \tt \int {\dfrac{2x.sin^{-1}2x}{\sqrt{1-4x^2}}} \, dx=\dfrac{1}{2}[2x-sin^{-1}2x\sqrt{1-4x^2}\;]$

$\bf \implies \tt \int^{\frac{\sqrt{3}}{4}}_0 \dfrac{2xsin^{-1}2x}{\sqrt{1-4x^2}}dx\\\\\implies \tt \dfrac{1}{2}\bigg[ 2x-sin^{-1}2x\sqrt{1-4x^2}\bigg]^{\frac{\sqrt{3}}{4}}_{0}\\\\\implies \tt \dfrac{1}{2}\bigg[ 2\bigg(\dfrac{\sqrt{3}}{4}\bigg) - sin^{-1}2 \bigg(\dfrac{\sqrt{3}}{4} \bigg) \times \sqrt{1-4\bigg( \dfrac{\sqrt{3}}{4}\bigg)^2}\bigg]-0\\\\\implies \tt \dfrac{1}{2}\bigg[ \dfrac{\sqrt{3}}{2}-sin^{-1}\bigg(\dfrac{\sqrt{3}}{2} \bigg) \times \dfrac{1}{2}\bigg]$

$\implies \tt \dfrac{1}{2}\bigg[\dfrac{\sqrt{3}}{2}-\dfrac{\pi}{3}\times \dfrac{1}{2}\bigg]\\\\\implies \tt \dfrac{1}{2}\bigg[\dfrac{\sqrt{3}}{2}-\dfrac{\pi}{6}\bigg]\\\\\implies \bf \bigg[ \dfrac{\sqrt{3}}{4}-\dfrac{\pi}{12} \bigg]$

Answer 2

Answer:

Answer

√3/4 - π/12

Given

$\bullet \;\;\bf \int\limits^{\frac{\sqrt{3}}{4}}_0 {\dfrac{2x.sin^{-1}2x}{\sqrt{1-4x^2}}}$\, dx∙

0

∫

4

3

1−4x

2

2x.sin

−1

2x

dx

To Find

Solve the integral

Solution

$\begin{lgathered}\tt \int {\dfrac{2x.sin^{-1}2x}{\sqrt{1-4x^2}}} \, dx\\\\\tt Let,t=sin^{-1}2x\;\; \& \;\; sint=2x\\\\\implies dt=\dfrac{2}{\sqrt{1-4x^2}}dx\;\; \& \;\; x=\dfrac{sint}{2}\end{lgathered}$

∫

1−4x

2

2x.sin

−1

2x

dx

Let,t=sin

−1

2x&sint=2x

⟹dt=

1−4x

2

2

dx&x=

2

sint

$\begin{lgathered}\tt \int \dfrac{sint}{2}tdt\\\\\implies \sf \dfrac{1}{2}\int t.sintdt ...(1)[\; Apply\ ILATE\ rule \;]\end{lgathered}$

∫

2

sint

tdt

⟹

2

1

∫t.sintdt...(1)[Apply ILATE rule]

Then ,

$\begin{lgathered}\implies \tt u=t\;\;\&\;\;dv=sintdt\\\\\implies \tt du=dt \;\; \& \;\; v=-cost\end{lgathered}$

⟹u=t&dv=sintdt

⟹du=dt&v=−cost

A/c to ILATE rule ,

uv - ∫ vdu

⇒ - t cos t - ∫ - cos t dt

⇒ - t cos t + ∫ cos t dt

⇒ - t cos t + sin t

⇒ sin t - t cos t

Now sub. all of above values .

$\implies \tt \dfrac{1}{2}\big[2x-sin^{-1}x\sqrt{1-4x^2}\;\big](From\ (1))$⟹

2

1

$2x−sin </p><p>−1</p><p> x </p><p>1−4x </p><p>2</p><p> </p><p> </p><p>$(From (1))

So , finally we get ,

$\begin{lgathered}\bullet \;\; \tt \int {\dfrac{2x.sin^{-1}2x}{\sqrt{1-4x^2}}} \, dx=\dfrac{1}{2}[2x-sin^{-1}2x\sqrt{1-4x^2}\;]\\\\\end{lgathered}$

∙∫

1−4x

2

2x.sin

−1

2x

dx=

2

1

$2x−sin </p><p>−1</p><p> 2x </p><p>1−4x </p><p>2</p><p> </p><p> </p><p>$

$\begin{lgathered}\bf \implies \tt \int^{\frac{\sqrt{3}}{4}}_0 \dfrac{2xsin^{-1}2x}{\sqrt{1-4x^2}}dx\\\\\implies \tt \dfrac{1}{2}\bigg[ 2x-sin^{-1}2x\sqrt{1-4x^2}\bigg]^{\frac{\sqrt{3}}{4}}_{0}\\\\\implies \tt \dfrac{1}{2}\bigg[ 2\bigg(\dfrac{\sqrt{3}}{4}\bigg) - sin^{-1}2 \bigg(\dfrac{\sqrt{3}}{4} \bigg) \times \sqrt{1-4\bigg( \dfrac{\sqrt{3}}{4}\bigg)^2}\bigg]-0\\\\\implies \tt \dfrac{1}{2}\bigg[ \dfrac{\sqrt{3}}{2}-sin^{-1}\bigg(\dfrac{\sqrt{3}}{2} \bigg) \times \dfrac{1}{2}\bigg]\\\\\end{lgathered}$

⟹∫

0

4

3

1−4x

2

2xsin

−1

2x

dx

⟹

2

1

$2x−sin </p><p>−1</p><p> 2x </p><p>1−4x </p><p>2</p><p> </p><p> </p><p>$

0

4

3

⟹

2

1

$2( </p><p>4</p><p>3</p><p> </p><p> </p><p> </p><p> )−sin </p><p>−1</p><p> 2( </p><p>4</p><p>3</p><p> </p><p> </p><p> </p><p> )× </p><p>1−4( </p><p>4</p><p>3</p><p> </p><p> </p><p> </p><p> ) </p><p>2</p><p> </p><p> </p><p>$−0

⟹

2

1

$</p><p>2</p><p>3</p><p> </p><p> </p><p> </p><p> −sin </p><p>−1</p><p> ( </p><p>2</p><p>3</p><p> </p><p> </p><p> </p><p> )× </p><p>2</p><p>1</p><p> </p><p>$

$\begin{lgathered}\implies \tt \dfrac{1}{2}\bigg[\dfrac{\sqrt{3}}{2}-\dfrac{\pi}{3}\times \dfrac{1}{2}\bigg]\\\\\implies \tt \dfrac{1}{2}\bigg[\dfrac{\sqrt{3}}{2}-\dfrac{\pi}{6}\bigg]\\\\\implies \bf \bigg[ \dfrac{\sqrt{3}}{4}-\dfrac{\pi}{12} \bigg]\end{lgathered}$

⟹

2

1

2

3

−

3

π

×

2

1

2

1

=

2

3

−

6

π

=

4

3

−

12

π