A Non-Trivial Integration by Parts
Z
arcsin2 x dx
Without knowing the anti-derivative of arcsin x the only choice is integration by parts with
u =
du =
Thus
arcsin2 x. 2 arcsin x
√ 1
1−x2
dx.
dv
= dx.
v
= x.
1
dx.
1 − x2
Now, recognizing that arcsin x and its derivative are in the resulting expression, we substitute as follows.
Z
arcsin2 x dx = x arcsin2 x −
θ
dθ
x
Now we have
Z
Z
x2 arcsin x
√
= arcsin x.
1
= √1−x
.
2
= sin θ
2
Z
2
arcsin x dx = x arcsin x − 2
sin θ(θ) dθ.
Now we solve the simpler integration by parts problem using
u = θ.
du = dθ.
dv
v
= sin θ dθ.
= − cos θ.
Z
Z
θ sin θ dθ
=
−θ cos θ −
− cos θ dθ
=
−θ cos θ + sin θ + C.
Note
√ if sin θ = x then the√opposite is x and the hypotenuse is 1. By the Pythagoran theorem the adjacent
is 1 − x2 . Thus cos θ = 1 − x2 . Returning to the original problem.
Z
arcsin2 x dx =
=
p
x arcsin2 x − 2 − arcsin x 1 − x2 + x + C
p
x arcsin2 x + 2 1 − x2 arcsin x − 2x + C.
An alternate solution uses the following.
u = arcsin x.
1
du = √1−x
dx
2
dv
v
Z
Z
arcsin x = x arcsin x −
=
=
√
dx.
x.
x
dx.
1 − x2
The remaining integral can be solved using the substitution u = 1 − x2 and du = −2x dx.
Z
Z
x
√
arcsin x = x arcsin x −
dx
1 − x2
Z
1
−2x
√
= x arcsin x +
dx
2
1 − x2
Z
1
= x arcsin x +
u−1/2 du
2
=
=
Now we try
R
arcsin2 x dx using parts as follows.
u =
du =
Z
x arcsin x + u1/2 + C
p
x arcsin x + 1 − x2 + C.
arcsin x.
√ 1
dx.
1−x2
dv
v
= arcsin x dx. √
= x arcsin x + 1 − x2 .
Z
p
p
1
2
√
arcsin x dx = arcsin x(x arcsin x + 1 − x ) −
x arcsin x + 1 − x2 dx
1 − x2
Z
p
x arcsin x
√
= arcsin x(x arcsin x + 1 − x2 ) −
+ 1 dx
1 − x2
Z
p
x arcsin x
√
= x arcsin2 x + arcsin x 1 − x2 − x −
dx
1 − x2
p
p
= x arcsin2 x + arcsin x 1 − x2 − x − − arcsin x 1 − x2 + x + C
p
= x arcsin2 x + 2 arcsin x 1 − x2 − 2x + C.
2
This used the same substitution (θ = arcsin x) as the first solution.