Let f(t) = ∫ ( (1 − sin(logₑ t)) / (1 − cos(logₑ t)) ) dt , t > 1. If f(e^{π/2}) = −e^{π/2} and f(e^{π/4}) = α e^{π/4}, then α equals

Let f(t) = ∫ ( (1 − sin(logₑ t)) / (1 − cos(logₑ t)) ) dt, t > 1. If f(e^{π/2}) = −e^{π/2} and f(e^{π/4}) = α e^{π/4}, then α equals

Q. Let $$ f(t)=\int\left(\frac{1-\sin(\log_e t)}{1-\cos(\log_e t)}\right)dt,\quad t>1. $$ If $$ f(e^{\pi/2})=-e^{\pi/2} $$ and $$ f(e^{\pi/4})=\alpha e^{\pi/4}, $$ then $\alpha$ equals

(A) $1+\sqrt{2}$

(B) $-1-2\sqrt{2}$

(D) $-1+\sqrt{2}$

Correct Answer: $-1-\sqrt{2}$

Explanation

Let

$$ x=\log_e t \;\Rightarrow\; t=e^x,\quad dt=e^x\,dx $$

Then,

$$ f(t)=\int e^x\frac{1-\sin x}{1-\cos x}\,dx $$

Use standard identities:

$$ 1-\cos x=2\sin^2\left(\frac{x}{2}\right),\quad \sin x=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) $$

So,

$$ \frac{1-\sin x}{1-\cos x} =\frac{1}{2}\csc^2\left(\frac{x}{2}\right)-\cot\left(\frac{x}{2}\right) $$

Hence,

$$ f(t)=\int e^x\left[\frac12\csc^2\left(\frac{x}{2}\right)-\cot\left(\frac{x}{2}\right)\right]dx $$

Put $u=\dfrac{x}{2}$, then $dx=2\,du$:

$$ f(t)=\int e^{2u}\left[\csc^2 u-2\cot u\right]du $$

Recognize:

$$ \frac{d}{du}\big(e^{2u}\cot u\big) =e^{2u}(2\cot u-\csc^2 u) $$

Therefore,

$$ f(t)=-e^{2u}\cot u+C=-e^x\cot\left(\frac{x}{2}\right)+C $$

Thus,

$$ f(t)=-t\cot\left(\frac{\log t}{2}\right)+C $$

Using $f(e^{\pi/2})=-e^{\pi/2}$:

$$ -e^{\pi/2}=-e^{\pi/2}\cot\left(\frac{\pi}{4}\right)+C $$

Since $\cot(\pi/4)=1$, we get $C=0$.

Now,

$$ f(e^{\pi/4})=-e^{\pi/4}\cot\left(\frac{\pi}{8}\right) $$

Using $\cot(\pi/8)=1+\sqrt{2}$:

$$ f(e^{\pi/4})=-(1+\sqrt{2})e^{\pi/4} $$

Hence,

$$ \alpha=\boxed{-1-\sqrt{2}} $$

Explanation

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