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Joint Entrance Examination
Mathematics
Indefinite Integrals
Question 304
Question 304 of 416
Medium
MCQ
The integral
∫
sin
2
x
cos
2
x
(
sin
3
x
+
cos
3
x
)
2
d
x
\displaystyle\int \dfrac{\sin^2x\cos^2x}{(\sin^3x+\cos^3x)^2}\,dx
∫
(
sin
3
x
+
cos
3
x
)
2
sin
2
x
cos
2
x
d
x
is equal to:
A
1
(
1
+
cot
3
x
)
+
c
\dfrac{1}{(1+\cot^3x)}+c
(
1
+
cot
3
x
)
1
+
c
B
−
1
3
(
1
+
tan
3
x
)
+
c
\dfrac{-1}{3(1+\tan^3x)}+c
3
(
1
+
tan
3
x
)
−
1
+
c
C
sin
3
x
(
1
+
cos
3
x
)
+
c
\dfrac{\sin^3x}{(1+\cos^3x)}+c
(
1
+
cos
3
x
)
sin
3
x
+
c
D
−
cos
3
x
3
(
1
+
sin
3
x
)
+
c
\dfrac{-\cos^3x}{3(1+\sin^3x)}+c
3
(
1
+
sin
3
x
)
−
cos
3
x
+
c
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Solutions · 1
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Official solution
I
=
∫
sin
x
cos
x
cos
2
x
(
1
+
tan
3
x
)
2
d
x
=
∫
sin
x
sec
2
x
(
1
+
tan
3
x
)
2
d
x
I = \int \dfrac{\sin x\cos x}{\cos^2x(1+\tan^3x)^2}\,dx = \int \dfrac{\sin x\sec^2x}{(1+\tan^3x)^2}\,dx
I
=
∫
cos
2
x
(
1
+
tan
3
x
)
2
sin
x
cos
x
d
x
=
∫
(
1
+
tan
3
x
)
2
sin
x
sec
2
x
d
x
Put
1
+
tan
3
x
=
t
⇒
d
t
=
3
tan
2
x
sec
2
x
d
x
1+\tan^3x=t \Rightarrow dt = 3\tan^2x\sec^2x\,dx
1
+
tan
3
x
=
t
⇒
d
t
=
3
tan
2
x
sec
2
x
d
x
Working through the substitution:
I
=
1
3
∫
d
t
t
2
=
−
1
3
t
+
c
=
−
1
3
(
1
+
tan
3
x
)
+
c
I = \dfrac13\int \dfrac{dt}{t^2} = -\dfrac{1}{3t}+c = \dfrac{-1}{3(1+\tan^3x)}+c
I
=
3
1
∫
t
2
d
t
=
−
3
t
1
+
c
=
3
(
1
+
tan
3
x
)
−
1
+
c
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