Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 \tan x(\cos x - y) . If the curve passes through the point ( \pi \over 4,0 ) , then the value of \int\limits_0^\pi /2 y\,dx is equal to :

(2 - \sqrt 2 ) + \pi \over \sqrt 2

(2 + \sqrt 2 ) + \pi \over \sqrt 2

Differential Equations PYQ — JEE Main 2022 · 28 Jun · Shift 2 · Q32 with Solution

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Correct answer: B

Form the differential equation

The slope is given by $\frac{dy}{dx}=2\tan x(\cos x-y).$

Rewrite it in linear form: $\frac{dy}{dx}+2\tan x\, y=2\tan x\cos x=2\sin x.$

So the differential equation is $\frac{dy}{dx}+2\tan x\,y=2\sin x.$

Solve the linear differential equation

Here, $P(x)=2\tan x, \qquad Q(x)=2\sin x.$

The integrating factor is $\text{I.F.}=e^{\int 2\tan x\,dx}.$

Since $\int \tan x\,dx=-\ln(\cos x),$ we get $\text{I.F.}=e^{-2\ln(\cos x)}=\sec^2 x.$

Multiply the equation by $\sec^2 x$ : $\sec^2 x\frac{dy}{dx}+2\tan x\sec^2 x\,y=2\sin x\sec^2 x.$

The left side becomes $\frac{d}{dx}(y\sec^2 x).$

Hence, $\frac{d}{dx}(y\sec^2 x)=2\sin x\sec^2 x=2\tan x\sec x.$

Integrating, $y\sec^2 x=\int 2\tan x\sec x\,dx=2\sec x+C.$

Therefore, $y=(2\sec x+C)\cos^2 x=2\cos x+C\cos^2 x.$

Use the given point

The curve passes through $\left(\frac{\pi}{4},0\right).$

Substitute $x=\frac{\pi}{4}$ and $y=0$ : $0=2\cos\frac{\pi}{4}+C\cos^2\frac{\pi}{4}.$

Now, $\cos\frac{\pi}{4}=\frac{1}{\sqrt2}, \qquad \cos^2\frac{\pi}{4}=\frac12.$

So,

=\sqrt2+\frac{C}{2}.$$ Thus, $$C=-2\sqrt2.$$ Hence the curve is $$y=2\cos x-2\sqrt2\cos^2 x.$$ --- 4. **Evaluate the integral** We need $$\int_0^{\pi/2} y\,dx=\int_0^{\pi/2}\left(2\cos x-2\sqrt2\cos^2 x\right)dx.$$ Split it: $$\int_0^{\pi/2} y\,dx=2\int_0^{\pi/2}\cos x\,dx-2\sqrt2\int_0^{\pi/2}\cos^2 x\,dx.$$ Now, $$\int_0^{\pi/2}\cos x\,dx=[\sin x]_0^{\pi/2}=1,$$ so $$2\int_0^{\pi/2}\cos x\,dx=2.$$ Also, $$\int_0^{\pi/2}\cos^2 x\,dx=\frac{\pi}{4}.$$ Therefore, $$\int_0^{\pi/2} y\,dx=2-2\sqrt2\cdot \frac{\pi}{4}=2-\frac{\pi}{\sqrt2}.$$ --- 5. **Match with the options** $$2-\frac{\pi}{\sqrt2}$$ corresponds to **Option B**. --- 6. **Comparison with stored correct answer** Stored correct answer: **B** Our derived answer: **B** So they agree.

More from Differential Equations

View written solutionFree

Correct answer: B

Form the differential equation

The slope is given by $\frac{dy}{dx}=2\tan x(\cos x-y).$

Rewrite it in linear form: $\frac{dy}{dx}+2\tan x\, y=2\tan x\cos x=2\sin x.$

So the differential equation is $\frac{dy}{dx}+2\tan x\,y=2\sin x.$

Solve the linear differential equation

Here, $P(x)=2\tan x, \qquad Q(x)=2\sin x.$

The integrating factor is $\text{I.F.}=e^{\int 2\tan x\,dx}.$

Since $\int \tan x\,dx=-\ln(\cos x),$ we get $\text{I.F.}=e^{-2\ln(\cos x)}=\sec^2 x.$

Multiply the equation by $\sec^2 x$ : $\sec^2 x\frac{dy}{dx}+2\tan x\sec^2 x\,y=2\sin x\sec^2 x.$

The left side becomes $\frac{d}{dx}(y\sec^2 x).$

Hence, $\frac{d}{dx}(y\sec^2 x)=2\sin x\sec^2 x=2\tan x\sec x.$

Integrating, $y\sec^2 x=\int 2\tan x\sec x\,dx=2\sec x+C.$

Therefore, $y=(2\sec x+C)\cos^2 x=2\cos x+C\cos^2 x.$

Use the given point

The curve passes through $\left(\frac{\pi}{4},0\right).$

Substitute $x=\frac{\pi}{4}$ and $y=0$ : $0=2\cos\frac{\pi}{4}+C\cos^2\frac{\pi}{4}.$

Now, $\cos\frac{\pi}{4}=\frac{1}{\sqrt2}, \qquad \cos^2\frac{\pi}{4}=\frac12.$

So,

=\sqrt2+\frac{C}{2}.$$ Thus, $$C=-2\sqrt2.$$ Hence the curve is $$y=2\cos x-2\sqrt2\cos^2 x.$$ --- 4. **Evaluate the integral** We need $$\int_0^{\pi/2} y\,dx=\int_0^{\pi/2}\left(2\cos x-2\sqrt2\cos^2 x\right)dx.$$ Split it: $$\int_0^{\pi/2} y\,dx=2\int_0^{\pi/2}\cos x\,dx-2\sqrt2\int_0^{\pi/2}\cos^2 x\,dx.$$ Now, $$\int_0^{\pi/2}\cos x\,dx=[\sin x]_0^{\pi/2}=1,$$ so $$2\int_0^{\pi/2}\cos x\,dx=2.$$ Also, $$\int_0^{\pi/2}\cos^2 x\,dx=\frac{\pi}{4}.$$ Therefore, $$\int_0^{\pi/2} y\,dx=2-2\sqrt2\cdot \frac{\pi}{4}=2-\frac{\pi}{\sqrt2}.$$ --- 5. **Match with the options** $$2-\frac{\pi}{\sqrt2}$$ corresponds to **Option B**. --- 6. **Comparison with stored correct answer** Stored correct answer: **B** Our derived answer: **B** So they agree.

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