If the integral 525 \int\limits_0^\frac\pi2 \sin 2x \cos^\frac112 x(1+\cos^\frac52 x)^\frac12 dx is equal to (n \sqrt2-64) , then n is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2024 · 31 Jan · Shift 1 · Q54 with Solution

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Correct answer: 176

Given integral

We need to evaluate

I=525\int_0^{\pi/2} \sin 2x\,\cos^{11/2}x\,\left(1+\cos^{5/2}x\right)^{1/2}dx.

We are told that

I=n\sqrt2-64.

Use $\sin 2x=2\sin x\cos x$

So,

I=525\int_0^{\pi/2}2\sin x\cos x\cdot \cos^{11/2}x\cdot \left(1+\cos^{5/2}x\right)^{1/2}dx.

Hence,

I=1050\int_0^{\pi/2}\sin x\cos^{13/2}x\left(1+\cos^{5/2}x\right)^{1/2}dx.

Substitute Let

t=1+\cos^{5/2}x.

Then

\frac{dt}{dx}=\frac{5}{2}\cos^{3/2}x(-\sin x)=-\frac52\sin x\cos^{3/2}x.

So,

sin x\cos^{3/2}x\,dx=-\frac{2}{5}dt.

Now write

\sin x\cos^{13/2}x\,dx na=\cos^5x\left(\sin x\cos^{3/2}x\,dx\right).

Since

\cos^5x=(\cos^{5/2}x)^2=(t-1)^2,

we get

I=1050\int_0^{\pi/2}(t-1)^2 t^{1/2}\left(-\frac25 dt\right).

Change limits

When $x=0$ ,

\cos x=1 \implies t=1+1=2.

When $x=\pi/2$ ,

\cos x=0 \implies t=1.

Therefore,

I=1050\left(-\frac25\right)\int_2^1 (t-1)^2 t^{1/2}dt =420\int_1^2 (t-1)^2 t^{1/2}dt.

Expand:

(t-1)^2=t^2-2t+1.

Thus,

I=420\int_1^2 \left(t^{5/2}-2t^{3/2}+t^{1/2}\right)dt.

Integrate

\int t^{5/2}dt=\frac{2}{7}t^{7/2}, \qquad \int t^{3/2}dt=\frac{2}{5}t^{5/2}, \qquad \int t^{1/2}dt=\frac{2}{3}t^{3/2}.

So,

I=420\left[\frac{2}{7}t^{7/2}-2\cdot\frac{2}{5}t^{5/2}+\frac{2}{3}t^{3/2}\right]_1^2

=420\left[\frac{2}{7}t^{7/2}-\frac{4}{5}t^{5/2}+\frac{2}{3}t^{3/2}\right]_1^2.

Evaluate at the limits

At $t=2$ :

t^{3/2}=2\sqrt2, \quad t^{5/2}=4\sqrt2, \quad t^{7/2}=8\sqrt2.

Hence,

\frac{2}{7}(8\sqrt2)-\frac{4}{5}(4\sqrt2)+\frac{2}{3}(2\sqrt2) =\left(\frac{16}{7}-\frac{16}{5}+\frac{4}{3}\right)\sqrt2.

Taking LCM $105$ ,

\frac{240-336+140}{105}\sqrt2=\frac{44}{105}\sqrt2.

At $t=1$ :

\frac{2}{7}-\frac{4}{5}+\frac{2}{3} =\frac{30-84+70}{105}=\frac{16}{105}.

Therefore,

I=420\left(\frac{44}{105}\sqrt2-\frac{16}{105}\right).

Since

\frac{420}{105}=4,

we get

I=4(44\sqrt2-16)=176\sqrt2-64.

Compare with $n\sqrt2-64$

Thus,

n\sqrt2-64=176\sqrt2-64.

So,

\boxed{n=176}.

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View written solutionFree

Correct answer: 176

Given integral

We need to evaluate

I=525\int_0^{\pi/2} \sin 2x\,\cos^{11/2}x\,\left(1+\cos^{5/2}x\right)^{1/2}dx.

We are told that

I=n\sqrt2-64.

Use $\sin 2x=2\sin x\cos x$

So,

I=525\int_0^{\pi/2}2\sin x\cos x\cdot \cos^{11/2}x\cdot \left(1+\cos^{5/2}x\right)^{1/2}dx.

Hence,

I=1050\int_0^{\pi/2}\sin x\cos^{13/2}x\left(1+\cos^{5/2}x\right)^{1/2}dx.

Substitute Let

t=1+\cos^{5/2}x.

Then

\frac{dt}{dx}=\frac{5}{2}\cos^{3/2}x(-\sin x)=-\frac52\sin x\cos^{3/2}x.

So,

sin x\cos^{3/2}x\,dx=-\frac{2}{5}dt.

Now write

\sin x\cos^{13/2}x\,dx na=\cos^5x\left(\sin x\cos^{3/2}x\,dx\right).

Since

\cos^5x=(\cos^{5/2}x)^2=(t-1)^2,

we get

I=1050\int_0^{\pi/2}(t-1)^2 t^{1/2}\left(-\frac25 dt\right).

Change limits

When $x=0$ ,

\cos x=1 \implies t=1+1=2.

When $x=\pi/2$ ,

\cos x=0 \implies t=1.

Therefore,

I=1050\left(-\frac25\right)\int_2^1 (t-1)^2 t^{1/2}dt =420\int_1^2 (t-1)^2 t^{1/2}dt.

Expand:

(t-1)^2=t^2-2t+1.

Thus,

I=420\int_1^2 \left(t^{5/2}-2t^{3/2}+t^{1/2}\right)dt.

Integrate

\int t^{5/2}dt=\frac{2}{7}t^{7/2}, \qquad \int t^{3/2}dt=\frac{2}{5}t^{5/2}, \qquad \int t^{1/2}dt=\frac{2}{3}t^{3/2}.

So,

I=420\left[\frac{2}{7}t^{7/2}-2\cdot\frac{2}{5}t^{5/2}+\frac{2}{3}t^{3/2}\right]_1^2

=420\left[\frac{2}{7}t^{7/2}-\frac{4}{5}t^{5/2}+\frac{2}{3}t^{3/2}\right]_1^2.

Evaluate at the limits

At $t=2$ :

t^{3/2}=2\sqrt2, \quad t^{5/2}=4\sqrt2, \quad t^{7/2}=8\sqrt2.

Hence,

\frac{2}{7}(8\sqrt2)-\frac{4}{5}(4\sqrt2)+\frac{2}{3}(2\sqrt2) =\left(\frac{16}{7}-\frac{16}{5}+\frac{4}{3}\right)\sqrt2.

Taking LCM $105$ ,

\frac{240-336+140}{105}\sqrt2=\frac{44}{105}\sqrt2.

At $t=1$ :

\frac{2}{7}-\frac{4}{5}+\frac{2}{3} =\frac{30-84+70}{105}=\frac{16}{105}.

Therefore,

I=420\left(\frac{44}{105}\sqrt2-\frac{16}{105}\right).

Since

\frac{420}{105}=4,

we get

I=4(44\sqrt2-16)=176\sqrt2-64.

Compare with $n\sqrt2-64$

Thus,

n\sqrt2-64=176\sqrt2-64.

So,

\boxed{n=176}.

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