250+ TOP MCQs on Methods of Integration-2 | Class 12 Maths

Mathematics MCQs for Engineering Entrance Exams on “Methods of Integration-2”.

1. Integrate 2 sin²⁡x+cos²x.
a) (frac{3x}{2}+frac{sin⁡2x}{4}+C)
b) (frac{3x}{2}-frac{sin⁡2x}{4}+C)
c) (frac{x}{2}+frac{sin⁡2x}{4}+C)
d) (frac{3x}{4}-frac{2sin⁡2x}{2}+C)
Answer: b
Clarification: (int ,2 ,sin^2⁡x +cos^2⁡x=int sin^2⁡x+sin^2⁡x+cos^2⁡x dx)
=(int sin^2⁡x+1 dx)
=(int sin^2⁡x dx+int 1 dx)
=(int frac{1-cos⁡2x}{2} dx+int 1 dx)
=(frac{x}{2}-frac{sin⁡2x}{4}+x)
=(frac{3x}{2}-frac{sin⁡2x}{4}+C)

2. Integrate 8 tan^3⁡x sec^2⁡x.
a) 2 tan⁴⁡x+C
b) 4 cot⁴⁡x+C
c) 2 tan^3⁡x+C
d) tan^4⁡x+C
Answer: a
Clarification: To find: (int 8 ,tan^3⁡x ,sec^2⁡x ,dx)
Let tan⁡x=t
sec^2⁡x dx=dt
∴(int 8 ,tan^3⁡x ,sec^2⁡x ,dx=int 8 ,t^3 ,dt=frac{8t^4}{4}=2t^4)
Replacing t with tan⁡x, we get
(int 8 tan^3⁡x sec^2⁡x dx=2 tan^4⁡x+C)

3. Find the integral of (frac{cos^2⁡x-sin^2⁡x}{7 cos^2⁡x sin^2⁡x}).
a) –(frac{1}{7}) (cot⁡x-tan⁡x)+C
b) –(frac{1}{7}) (cot⁡x-2 tan⁡x)+C
c) –(frac{1}{7}) (cot⁡x+tan⁡x)+C
d) –(frac{1}{7}) (2 cot⁡x+3 tan⁡x)+C
Answer: c
Clarification: To find: (int frac{cos^2⁡x-sin^2⁡x}{7 ,cos^2⁡x ,sin^2⁡x} dx)
(int frac{cos^2⁡x-sin^2⁡x}{7 ,cos^2⁡x ,sin^2⁡x} dx=frac{1}{7} int frac{1}{sin^2⁡x}-frac{1}{cos^2⁡x} dx)
=(frac{1}{7} int cosec^2 x-sec^2⁡x dx)
=(frac{1}{7}) (-cot⁡x-tan⁡x)+C
=-(frac{1}{7}) (cot⁡x+tan⁡x)+C.

4. Find (int sin^2⁡(8x+5) dx)
a) (frac{x}{4}+frac{sin⁡(16x+10)}{32}+C)
b) (frac{x}{2}-frac{cos⁡(16x+10)}{32}+C)
c) (frac{x}{2}-frac{sin⁡(16x+10)}{32}+C)
d) (frac{x}{2}+frac{cos⁡(16x+5)}{32}+C)
Answer: c
Clarification: (int sin^2⁡(8x+5) dx=int frac{1-cos⁡2(8x+5)}{2} dx=int frac{1}{2} dx-frac{1}{2} int cos(16x+10)dx)
=(frac{x}{2}-frac{1}{2} (frac{sin⁡(16x+10)}{16})=frac{x}{2}-frac{sin⁡(16x+10)}{32}+C)

5. Find (int frac{5 cos^2⁡x}{1+sin⁡x} dx).
a) -3(x+cos⁡x)+C
b) 5(x+cos⁡x)+C
c) 5(-x+sin⁡x)+C
d) 5(x-cos⁡x)+C
Answer: b
Clarification: (int frac{5 cos^2⁡x}{1+sin⁡x} dx=int frac{5(1-sin^2⁡x)}{1+sin⁡x}=5int frac{(1+sin⁡x)(1-sin⁡x)}{(1+sin⁡x)} dx)
=5∫ (1-sin⁡x)dx
=5(x-(-cos⁡x))=5(x+cos⁡x)+C

6. Find the integral of (frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})}).
a) cot⁡xe^-x+C
b) -cot⁡xe^-x+C
c) -cot⁡xe^x+C
d) -cos^2⁡xe^-x+C
Answer: b
Clarification: (int frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})} dx)
Let xe^-x=t
Differentiating w.r.t x, we get
(-xe^{-x}+e^{-x} dx=dt)
e^-x (1-x)dx=dt
(int frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})} dx=int frac{dt}{sin^2⁡t})
=(int cosec^2 ,t ,dt)
=-cot⁡t+C
Replacing t with xe^-x, we get
(int frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})} dx=-cot⁡xe^{-x}+C).

7. Integrate (frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2}).
a) -log⁡(1+2sin⁡2x)+C
b) (frac{1}{4}) log⁡(1-sin⁡2x)+C
c) –(frac{1}{4}) log⁡(1+cos⁡2x)+C
d) -log⁡(1-sin⁡2x)+C
Answer: d
Clarification: (int frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2} dx=int frac{2 cos⁡2x}{cos^2⁡x+sin^2⁡x-sin⁡2x} ,dx ,(∵2 cos⁡x sin⁡x=sin⁡2x))
=(int frac{2 cos⁡2x}{1-sin⁡2x} dx)
Let 1-sin⁡2x=t
Differentiating w.r.t x, we get
-2 cos⁡2x dx=dt
2 cos⁡2x dx=-dt
(int frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2} dx=-int frac{dt}{t})
=-log⁡t
Replacing t with 1-sin⁡2x, we get
∴(int frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2}) dx=-log⁡(1-sin⁡2x)+C

8. Integrate sin³⁡(x+2).
a) (frac{3}{4} ,(sin⁡(x+2))+frac{1}{12} ,cos⁡(3x+6)+C)
b) –(frac{3}{4} ,(cos⁡(x+2))-frac{1}{5} ,cos⁡(3x+6)+C)
c) –(frac{3}{4} ,(cos⁡(x+2))+frac{1}{12} ,cos⁡(3x+6)+C)
d) –(frac{3}{4} ,(cos⁡(x+2))+frac{1}{12} ,sin⁡(x+2)+C)
Answer: c
Clarification: To find: ∫ 3 sin³⁡(x+2) dx
We know that, sin⁡3x=3 sin⁡x-4 sin³⁡x
∴sin³⁡⁡x=(frac{3 sin⁡x-sin⁡3x}{4})
sin³⁡(x+2)=(frac{(3 sin⁡(x+2)-sin⁡(3x+6))}{4})
(int sin^3⁡(x+2) ,dx=frac{3}{4} int sin⁡(x+2) ,dx-frac{1}{4} int ,sin⁡(3x+6) ,dx)
=-(frac{3}{4} ,(cos⁡(x+2))+frac{1}{12} ,cos⁡(3x+6)+C)

9. Integrate 2x cos⁡(x²+3).
a) sin⁡(x²+3)+C
b) sin²⁡(x²+3)+C
c) cot⁡(x²+3)+C
d) -sin⁡(x²+3)+C
Answer: a
Clarification: ∫ 2x cos⁡(x²+3) dx
Let x²+3=t
Differentiating w.r.t x, we get
2x dx=dt
∫ 2x cos⁡(x²+3) dx=∫ cos⁡t dt
=sin⁡t+C
Replacing w.r.t x, we get
∴∫ 2x cos⁡(x²+3) dx=sin⁡(x²+3)+C

10. Find ∫ 2 sin³⁡x+1 dx
a) (frac{3}{2}-frac{cos⁡3x}{6}+x+C)
b) –(frac{3}{2} cos⁡x+frac{cos⁡3x}{6}+x+C)
c) –(frac{3}{2} cos⁡x-frac{cos⁡3x}{6}-x+C)
d) –(frac{3}{2} cos⁡x+frac{cos⁡3x}{6}+C)
Answer: b
Clarification: We know that, sin⁡3x=3 sin⁡x-4 sin³⁡x
∴sin³x=(frac{3 sin⁡x-sin⁡3x}{4})
(int 2 ,sin^3⁡x+1 ,dx=int frac{(3 sin⁡x-sin⁡3x)}{2} dx+int dx)
=(frac{3}{2} int sin⁡x dx-frac{1}{2} int sin⁡3x dx+int dx)
=-(frac{3}{2} cos⁡x+frac{cos⁡3x}{6}+x+C)

Mathematics MCQs for Engineering Entrance Exams,