微分積分学‎:
積分: 三角関数の置換積分

1.  
∫116+x2 dx\int \frac{1}{16+{x}^{2}} \, dx
  解答
2.  
∫3x2−9 dx\int \frac{3}{\sqrt{{x}^{2}-9}} \, dx
  解答
3.  
∫225−x2 dx\int \frac{2}{\sqrt{25-{x}^{2}}} \, dx
  解答
4.  
∫x2−16x dx\int \frac{\sqrt{{x}^{2}-16}}{x} \, dx
  解答
5.  
∫1(1+x2)2 dx\int \frac{1}{{(1+{x}^{2})}^{2}} \, dx
  解答
6.  
∫1x44−x2 dx\int \frac{1}{{x}^{4}\sqrt{4-{x}^{2}}} \, dx
  解答
7.  
∫25x2−4x dx\int \frac{\sqrt{25{x}^{2}-4}}{x} \, dx
  解答
8.  
∫1x2x2−9 dx\int \frac{1}{{x}^{2}\sqrt{{x}^{2}-9}} \, dx
  解答

Trigonometric Substitution - Introduction

This tutorial assumes that you are familiar with trigonometric identities, derivatives, integration of trigonometric functions, and integration by substitution.
After simpler methods of integration failed, we should consider trigonometric substitution. The requirement is that the function contains the form
a2−x2{a}^{2}-{x}^{2}
, the form
a2+x2{a}^{2}+{x}^{2}
, or the form
x2−a2{x}^{2}-{a}^{2}
. Some examples are:
1.
x2−25\sqrt{{x}^{2}-25}

2.
(4−y2)32{(4-{y}^{2})}^{\frac{3}{2}}

3.
9−h2\sqrt{9-{h}^{2}}

4.
116+x2\frac{1}{16+{x}^{2}}
In fact, there are clear rules on what substitutions to make:
1. If the function contains
a2−x2{a}^{2}-{x}^{2}
, let
x=asin⁡ux=a\sin{u}
.
2. If the function contains
a2+x2{a}^{2}+{x}^{2}
, let
x=atan⁡ux=a\tan{u}
.
3. If the function contains
x2−a2{x}^{2}-{a}^{2}
, let
x=asec⁡ux=a\sec{u}
.

Using Trigonometric Substitution

Let's try an example. Solve this integral:
∫116+x2 dx\int \frac{1}{16+{x}^{2}} \, dx
Since the denominator fits the form
a2+x2{a}^{2}+{x}^{2}
, let's use Rule #2 from the section above.
We let
x=atan⁡ux=a\tan{u}
, which is
4tan⁡u4\tan{u}
.
Then, to get
dxdx
, we find the derivative of
4tan⁡u4\tan{u}
, which is
4sec⁡2u4\sec^{2}u
.
Therefore, let
x=atan⁡ux=a\tan{u}
, and
dx=4sec⁡2udx=4\sec^{2}u
.
Substituting the variables from the above, we have:
∫116+(4tan⁡u)2×4sec⁡2u du\int \frac{1}{16+{(4\tan{u})}^{2}}\times 4\sec^{2}u \, du
After simplifying, we have:
∫14 du\int \frac{1}{4} \, du
By the Power Rule, the integral is:
u4\frac{u}{4}
From the earlier steps, we know that
x=atan⁡ux=a\tan{u}
, and solving for
uu
, we have:
u=tan⁡−1(14x)u=\tan^{-1}{(\frac{1}{4}x)}
The only step remaining is to substitute the value of
uu
back into the earlier integral, namely
u4\frac{u}{4}
. After the substitution, we have:
tan⁡−1(14x)4+C\frac{\tan^{-1}{(\frac{1}{4}x)}}{4}+C
We are done. We have successfully used trigonometric substitution to find the integral.

What's Next

Ready to dive deeper? You can try more practice problems at the top of this page to help you get more familiar with solving integral using trigonometric substitution. Want even more help? Sign up for Cymath Plus today. You can also download the Cymath app for iOS and Android to get step-by-step help on-the-go.
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