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CBSE
MATHEMATICS · Single correct · CBSE · Mathematics · Integrals · Integrals
Problem
∫
x
sin
2
x
d
x
=
\int x\sin^2 x\,dx =
∫
x
sin
2
x
d
x
=
(a)
x
2
4
+
x
4
sin
2
x
+
1
8
cos
2
x
+
c
\frac{x^2}{4} + \frac{x}{4}\sin 2x + \frac{1}{8}\cos 2x + c
4
x
2
+
4
x
sin
2
x
+
8
1
cos
2
x
+
c
(b)
x
2
4
−
x
4
sin
2
x
+
1
8
cos
2
x
+
c
\frac{x^2}{4} - \frac{x}{4}\sin 2x + \frac{1}{8}\cos 2x + c
4
x
2
−
4
x
sin
2
x
+
8
1
cos
2
x
+
c
(c)
x
2
4
+
x
4
sin
2
x
−
1
8
cos
2
x
+
c
\frac{x^2}{4} + \frac{x}{4}\sin 2x - \frac{1}{8}\cos 2x + c
4
x
2
+
4
x
sin
2
x
−
8
1
cos
2
x
+
c
(d)
x
2
4
−
x
4
sin
2
x
−
1
8
cos
2
x
+
c
\frac{x^2}{4} - \frac{x}{4}\sin 2x - \frac{1}{8}\cos 2x + c
4
x
2
−
4
x
sin
2
x
−
8
1
cos
2
x
+
c
Related theory
Integration Techniques
Integration by Parts
Integrals
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