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

Put 1 + x³ = t²

⇒ 3x² dx = 2tdt and x³ = t² − 1

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

= [2 / 3] ∫{[(t² − 1) * t] dt / [t]}

= [2 / 3] ∫(t² − 1) dt

= [2 / 3] [(t³ / 3) − t] + c

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

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 x² + c

∫√(1 + sin [x / 2]) dx = ∫√(sin² [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

∫[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

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

∫(log x)² dx

Put log x = t

⇒ e^t = x

⇒ dx = e^t dt, then it reduces to

∫t² * e^t dt = t² * e^t − 2t * e^t + 2e^t + c

= x (log x)² − 2x log x + 2x + c

Put a + bx = t

⇒ x = [t − a] / [b] and dx = dt / [b]

I =∫([t − a] / b)² * [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)]

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

∫{[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

∫tan³2x sec²x dx = ∫[(sec² 2x − 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

= [sec³ 2x / 6] − [sec2x / 2] + c

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

[1 / 2] ∫(t² − 1) dt = [1 / 6]t³ − [1 / 2]t + c

= [sec³ 2x / 6] − [sec 2x / 2] + c