Indefinite IntegrationeasyFree

f(π/2) when f'(x)=1/(1+cosx), f(0)=3 | JEE

JEE Maths question with a full step-by-step solution.

Question
Let ff be continuous with f′(x)=11+cos⁡xf'(x)=\dfrac{1}{1+\cos x} and f(0)=3f(0)=3. Then f ⁣(π2)f\!\left(\dfrac{\pi}{2}\right) equals
A134\dfrac{13}{4}
B22
C44correct
DNone of these
Solution
Step 1: Integrate f′(x)f'(x). Multiply numerator and denominator by (1−cos⁡x)(1-\cos x):
11+cos⁡x=1−cos⁡x(1+cos⁡x)(1−cos⁡x)=1−cos⁡x1−cos⁡2x=1−cos⁡xsin⁡2x\dfrac{1}{1+\cos x}=\dfrac{1-\cos x}{(1+\cos x)(1-\cos x)}=\dfrac{1-\cos x}{1-\cos^{2}x}=\dfrac{1-\cos x}{\sin^{2}x}
 Step 2: Split:
1−cos⁡xsin⁡2x=1sin⁡2x−cos⁡xsin⁡2x=csc⁡2x−csc⁡xcot⁡x\dfrac{1-\cos x}{\sin^{2}x}=\dfrac{1}{\sin^{2}x}-\dfrac{\cos x}{\sin^{2}x}=\csc^{2}x-\csc x\cot x
 Step 3: Integrate:
f(x)=∫(csc⁡2x−csc⁡xcot⁡x) dx=−cot⁡x+csc⁡x+C=1−cos⁡xsin⁡x+C=tan⁡x2+Cf(x)=\int(\csc^{2}x-\csc x\cot x)\,dx=-\cot x+\csc x+C=\dfrac{1-\cos x}{\sin x}+C=\tan\dfrac{x}{2}+C
 Step 4: Apply f(0)=3f(0)=3: tan⁡0+C=C=3\tan 0+C=C=3. Step 5: Evaluate at x=π2x=\dfrac{\pi}{2}:
f ⁣(π2)=tan⁡π4+3=1+3=4f\!\left(\dfrac{\pi}{2}\right)=\tan\dfrac{\pi}{4}+3=1+3=4
 Correct answer: (3)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Indefinite Integration · medium
If ∫ex 1−x6−6x2(1+x3)2 dx=ex 1−xp1+xq+c\displaystyle\int e^{x}\,\dfrac{1-x^{6}-6x^{2}}{(1+x^{3})^{2}}\,dx=e^{x}\,\dfrac{1-x^{p}}{1+x^{q}}+c, then p+qp+q equals
Indefinite Integration · medium
∫dxx−x3\displaystyle\int \dfrac{dx}{\sqrt{x}-\sqrt[3]{x}} equals
Indefinite Integration · easy
If ∫x−sin⁡xcos⁡xx2cos⁡2x dx=f(x)+c\displaystyle\int \dfrac{x-\sin x\cos x}{x^{2}\cos^{2}x}\,dx=f(x)+c, then lim⁡x→0f(x)\displaystyle\lim_{x\to 0}f(x) equals
Indefinite Integration · medium
If ∫x2e−2x dx=e−2x(ax2+bx+c)+d\displaystyle\int x^{2}e^{-2x}\,dx=e^{-2x}(ax^{2}+bx+c)+d, then

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath