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Question: If $\int \frac{sin\theta - cos\theta}{(sin\theta + cos\theta)\sqrt{sin\theta cos\theta + sin^2\theta...

If ∫sinθ−cosθ(sinθ+cosθ)sinθcosθ+sin2θcos2θdθ=cosec−1(f(θ))+C\int \frac{sin\theta - cos\theta}{(sin\theta + cos\theta)\sqrt{sin\theta cos\theta + sin^2\theta cos^2\theta}}d\theta = cosec^{-1}(f(\theta)) + C, then

A

f(0)=sin2θ+1f(0) = sin2\theta + 1

B

f(0)=1−sin2θf(0) = 1 - sin2\theta

C

f(0)=sin2θ−1f(0) = sin2\theta - 1

D

none of these

Answer

f(0) = sin2θ + 1

Explanation

Solution

To solve the integral I=∫sin⁡θ−cos⁡θ(sin⁡θ+cos⁡θ)sin⁡θcos⁡θ+sin⁡2θcos⁡2θdθI = \int \frac{\sin\theta - \cos\theta}{(\sin\theta + \cos\theta)\sqrt{\sin\theta \cos\theta + \sin^2\theta \cos^2\theta}}d\theta, we use a substitution.

Let t=sin⁡θ+cos⁡θt = \sin\theta + \cos\theta.
Differentiating tt with respect to θ\theta:
dt=(cos⁡θ−sin⁡θ)dθdt = (\cos\theta - \sin\theta)d\theta
So, (sin⁡θ−cos⁡θ)dθ=−dt(\sin\theta - \cos\theta)d\theta = -dt.

Next, we express sin⁡θcos⁡θ\sin\theta\cos\theta in terms of tt:
t2=(sin⁡θ+cos⁡θ)2=sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θt^2 = (\sin\theta + \cos\theta)^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta
t2=1+2sin⁡θcos⁡θt^2 = 1 + 2\sin\theta\cos\theta
sin⁡θcos⁡θ=t2−12\sin\theta\cos\theta = \frac{t^2 - 1}{2}.

Now, substitute these into the expression under the square root in the denominator:
sin⁡θcos⁡θ+sin⁡2θcos⁡2θ=sin⁡θcos⁡θ(1+sin⁡θcos⁡θ)\sin\theta\cos\theta + \sin^2\theta\cos^2\theta = \sin\theta\cos\theta(1 + \sin\theta\cos\theta)
Substitute sin⁡θcos⁡θ=t2−12\sin\theta\cos\theta = \frac{t^2 - 1}{2}:
=t2−12(1+t2−12)= \frac{t^2 - 1}{2} \left(1 + \frac{t^2 - 1}{2}\right)
=t2−12(2+t2−12)= \frac{t^2 - 1}{2} \left(\frac{2 + t^2 - 1}{2}\right)
=t2−12(t2+12)= \frac{t^2 - 1}{2} \left(\frac{t^2 + 1}{2}\right)
=(t2−1)(t2+1)4= \frac{(t^2 - 1)(t^2 + 1)}{4}
=t4−14= \frac{t^4 - 1}{4}.

So the integral becomes:
I=∫−dttt4−14I = \int \frac{-dt}{t\sqrt{\frac{t^4 - 1}{4}}}
I=∫−dttt4−12I = \int \frac{-dt}{t \frac{\sqrt{t^4 - 1}}{2}}
I=∫−2dttt4−1I = \int \frac{-2dt}{t\sqrt{t^4 - 1}}.

To evaluate this integral, let u=t2u = t^2.
Then du=2tdtdu = 2t dt.
So, dt=du2tdt = \frac{du}{2t}.
Substitute uu and dtdt into the integral:
I=∫−2tu2−1du2tI = \int \frac{-2}{t\sqrt{u^2 - 1}} \frac{du}{2t}
I=∫−dut2u2−1I = \int \frac{-du}{t^2\sqrt{u^2 - 1}}
Since t2=ut^2 = u, we have:
I=∫−duuu2−1I = \int \frac{-du}{u\sqrt{u^2 - 1}}.

This is a standard integral form. We know that ∫dxxx2−a2=1asec⁡−1(xa)+C\int \frac{dx}{x\sqrt{x^2 - a^2}} = \frac{1}{a}\sec^{-1}\left(\frac{x}{a}\right) + C.
Here, a=1a=1.
So, I=−sec⁡−1(u)+CI = -\sec^{-1}(u) + C.

Now, substitute back u=t2u = t^2:
I=−sec⁡−1(t2)+CI = -\sec^{-1}(t^2) + C.

Finally, substitute back t=sin⁡θ+cos⁡θt = \sin\theta + \cos\theta:
t2=(sin⁡θ+cos⁡θ)2=sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=1+2sin⁡θcos⁡θ=1+sin⁡2θt^2 = (\sin\theta + \cos\theta)^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 1 + 2\sin\theta\cos\theta = 1 + \sin2\theta.
So, I=−sec⁡−1(1+sin⁡2θ)+CI = -\sec^{-1}(1 + \sin2\theta) + C.

The problem states that I=csc⁡−1(f(θ))+CI = \csc^{-1}(f(\theta)) + C.
We use the identity relating sec⁡−1\sec^{-1} and csc⁡−1\csc^{-1}: For x≥1x \ge 1, sec⁡−1(x)+csc⁡−1(x)=π2\sec^{-1}(x) + \csc^{-1}(x) = \frac{\pi}{2}.
Therefore, −sec⁡−1(x)=csc⁡−1(x)−π2-\sec^{-1}(x) = \csc^{-1}(x) - \frac{\pi}{2}.
Applying this to our integral:
I=csc⁡−1(1+sin⁡2θ)−π2+CI = \csc^{-1}(1 + \sin2\theta) - \frac{\pi}{2} + C.
Let C′=C−π2C' = C - \frac{\pi}{2}. Then:
I=csc⁡−1(1+sin⁡2θ)+C′I = \csc^{-1}(1 + \sin2\theta) + C'.

Comparing this with the given form I=csc⁡−1(f(θ))+CI = \csc^{-1}(f(\theta)) + C, we find:
f(θ)=1+sin⁡2θf(\theta) = 1 + \sin2\theta.

This matches option (1).