# Indefinite and Definite Integrals Past Year Solved Questions for IIT JEE & Engineering Exam

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## Indefinite and Definite Integrals Past Year Solved Questions

Indefinite and Definite Integrals Past Year Solved Questions: In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you Indefinite and Definite Integrals Past Year Solved Questions for IIT JEE & Engineering Exam.

## Indefinite and Definite Integrals Past Year Solved Questions

Q1. ∫[x5 / √(1 + x3)] dx = ________.

Put 1 + x= t2

⇒ 3xdx = 2tdt and x= t2 − 1

So, ∫[x5 / √(1 + x3)] dx = ∫{[x2 * x3] / √(1 + x3)} dx

= [2 / 3] ∫{[(t− 1) * t] dt / [t]}

= [2 / 3] ∫(t− 1) dt

= [2 / 3] [(t3 / 3) − t] + c

= [2 / 3] [{(1 + x3)3/2 / 3} − {(1 + x3)½}]+ c

Q2. ∫x / [1 + x4] dx = ________.

Put t = x2 ⇒ dt = 2x dx, therefore,

∫x / [1 + x4] dx = [1 / 2] ∫1 / [1 + t2] dt

= [1 / 2] tan−1 t + c

= [1 / 2] tan−1 x+ c

Q3. ∫√(1 + sin [x / 2]) dx = _________.

∫√(1 + sin [x / 2]) dx = ∫√(sin[x / 4] + cos2 [x / 4] + 2 sin [x / 4] cos [x / 4]) dx

= ∫(sin [x / 4] + cos [x / 4]) dx

= 4 (sin [x / 4] – cos [x / 4]) + c

Q4.  ∫[sinx] / [sin (x − α)] dx = ________

∫[sinx] / [sin (x − α)] dx =

∫[sin (x − α + α)] / [sin (x − α)] dx

= ∫{[(sin (x − α) cosα + cos (x − α) sinα] / [sin (x − α)]} dx

= ∫cosα dx +∫sinα * cot (x − α) dx

= x cosα + sinα * log sin (x − α) + c

Q5. ∫[1 + x2] / √[1 − x2] dx = ________.

Put x = sinθ ⇒ dx = cosθ dθ, then it reduces to

∫(1 + sin2θ) dθ = θ + [1 / 2]∫(1 − cos2θ) dθ

= [3θ / 2] − [1 / 2] sinθ * √[1 − sin2θ] + c

= [3 / 2] sin−1 x − [1 / 2]x √[1 − x2] + c

Q6. ∫(log x)2 dx = _______.

∫(log x)2 dx

Put log x = t

⇒ et = x

⇒ dx = et dt, then it reduces to

∫t2 * et dt = t* e− 2t * e+ 2et + c

= x (log x)− 2x log x + 2x + c

Q7. ∫x2dx / (a + bx)2 = ___________.

Put a + bx = t

⇒ x = [t − a] / [b] and dx = dt / [b]

I =∫([t − a] / b)* [1 / t2] * [dt / b]

= [1 / b2]∫(1 − (2a / t) + [a2 * t−2]) dt

= [1 / b2] * [(t − 2a log t) − (a2 / t)]

= [1 / b2] [(x + a / b) − [2a / b] * log (a + bx) − [a2 / b] * [1 / (a + bx)]

Q8. ∫x cos2x dx = ______.

x cos2x dx = [1 / 2] ∫x (1 + cos2x) dx

= [x2 / 4] + [1 / 2] [(x sin2x) / (2) −∫(sin2x / 2) dx] + c

= [x2 / 4] + (x sin2x / 4) + (cos2x / 8) + c

Q9. ∫{[sin8x − cos8x] / [1 − 2 sin2x cos2x]} dx = _________.

∫{[sin8x − cos8x] / [1 − 2 sin2x cos2x]} dx

= ∫{[(sin4x + cos4x) * (sin4x − cos4x)] / [(sin2x + cos2x)2 − 2 sin2x cos2x]} dx

= ∫(sin4x – cos4x) dx

= ∫[sin2x + cos2x] * [sin2x – cos2x] dx

= ∫(sin2x + cos2x) dx

= ∫−cos2xdx

= [−sin2x / 2] + c

Q10. ∫tan32x sec2x dx = __________.

∫tan32x sec2x dx = ∫[(sec2x − 1) sec2x * tan2x] dx

=∫sec32x tan2x dx −∫sec2x tan2x dx ……. (i)

Now, we take ∫sec32x tan2x dx

Put sec 2x = t

⇒ sec 2x tan 2x = dt/2, then it reduces to

[1 / 2] ∫t2 dt = t3 / 6

= [sec32x] / [6]

From (i), ∫sec32x tan2x dx −∫sec2x tan2x dx

= [sec2x / 6] − [sec2x / 2] + c

Trick: Let sec 2x = t, then sec 2x tan 2x dx = [1 / 2] dt

[1 / 2] ∫(t− 1) dt = [1 / 6]t− [1 / 2]t + c

= [sec2x / 6] − [sec 2x / 2] + c