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Calculus Integration Problems

This document contains solutions to 10 integration problems: 1) Finding the integral of x sin x and relating it to a constant term. 2) Integrating x tan-1 x and relating it to terms involving x and tan-1 x. 3) Solving the integral of e^-2x sin 3x using substitution and relating the solution to terms involving e^-2x, cos 3x, and sin 3x. 4) Integrating x^2 sin 2x using substitution and relating the solution to terms involving x, cos 2x, and sin 2x. 5) Integrating a term involving e^x, cos 2x, and sin 2x and relating

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PRADEEP C
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87 views2 pages

Calculus Integration Problems

This document contains solutions to 10 integration problems: 1) Finding the integral of x sin x and relating it to a constant term. 2) Integrating x tan-1 x and relating it to terms involving x and tan-1 x. 3) Solving the integral of e^-2x sin 3x using substitution and relating the solution to terms involving e^-2x, cos 3x, and sin 3x. 4) Integrating x^2 sin 2x using substitution and relating the solution to terms involving x, cos 2x, and sin 2x. 5) Integrating a term involving e^x, cos 2x, and sin 2x and relating

Uploaded by

PRADEEP C
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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1.

(a) Since,
∫ x sin x dx=−xcos x+ A
⇒−x cos x +sin x+constant=−x cos x+ A
Equating it, we get A=sin x+ constant .

2 2
+1−1
∫ x . tan−1 x dx= x2 tan−1 x− 12 ∫ x 1+ x2
dx
2. (a)
x 2 −1 1 1
= tan x− x + tan −1 x+ c
2 2 2
1 1
= tan−1 x . ( x 2 +1)− x +c
2 2 .

I=∫ e−2 x sin 3 x dx


3. (d) Let
e−2 x cos3 x 2e−2 x cos3 x
=− −∫ dx
3 3

=−
3
−
3 [
e−2 x cos3 x 2 e−2 x sin 3 x
3
+∫
2 e−2 x sin 3 x
3
dx ]
e−2 x cos3 x 2 e−2 x sin 3 x 4
⇒ I=− − − I
3 9 9

⇒
13
9
I=−e−2 x
[
3 cos 3 x +2 sin 3 x
9 ]
1 −2 x
I=− e [3 cos3 x +2 sin3 x ]
Hence 13 .

−x 2 cos2 x 2 x cos2 x
I=∫ x 2 sin 2 x dx= +∫ dx +c
4. (b) Let 2 2
x 2 cos2 x x sin 2 x cos 2 x
=− + + +c .
2 2 4

5. (d)
∫ ( 1+cos2 x)
2+sin 2 x x
e dx=∫ ( 2 ex
1+cos2 x )
dx +∫
e x sin 2 x
1+cos2 x
dx

=∫ e x sec 2 x dx+∫ e x tan x dx=e x tan x+c


.

1
sin−1 x =t ⇒ dx=dt ,
6. (a) Putting √1−x 2 we get
x sin−1 x
∫ dx= ∫ t sin t dt=−t cos t+sin t +c
√ 1−x 2
=−sin−1 x cos(sin−1 x )+sin(sin−1 x )+c
=x−sin−1 x √1−x 2 +c .

7. (c)
∫ √ x2−8 x+7 dx=∫ √( x−4)2−(3)2 dx
Now apply formula of
∫ √ x 2−a2 dx.

x 1
∫ √1+ x 2 dx= 2 √ x 2+1+ 2 log ( x +√ x2 +1)+c
8. (a) .

1
∫ e x sin x dx= 2 e x a+ c
9. (a) Given that ..…(i)

I=∫ e x sin x dx=−e x cos x+∫ e x cos x dx+c


Let

=−e x cos x+e x sin x−∫ e x sin x dx+c


⇒ 2 I=e x (−cos x +sin x )+c . Now from (i), we get
1 x 1
e a= e x (sin x−cos x )⇒ a=sin x−cos x .
2 2

I=∫ √ x 2 +a 2 dx =∫ √ x 2 +a2 .1 dx
10. (b)
= √ x 2 +a2 ∫ 1 dx−∫
[ d
dx
( √ x 2 +a2 ) ∫ 1 dx dx ]
=x √ x +a −∫
2 2
[√ 2 x +a
2x
2 2 ]
x dx

=x √ x2 +a2 − ∫[ x 2 +a 2−a2
dx
]
√ x 2 + a2

=x √ x2 +a2 − ∫ [√ x 2 +a 2−
a2
] dx
√ x 2+a 2
dx
=x √ x2 +a2 −∫ √ x 2 +a 2 dx+a2 ∫
√ x 2 +a2
= √ x 2+a2−I +a2 log [ x+√ x2 +a2 ] +C


2 I=x √ x 2+a 2+ a2 log [ x+√ x 2+a 2] +C
x 2 2 a2
I=
2
√ x + a + 2 log [ x + √ x 2 +a 2 ] +C
 .

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