Difference between revisions of "Science:Math Exam Resources/Courses/MATH105/April 2017/Question 01 (h)/Solution 1"

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Then using integration by parts we have  
 
Then using integration by parts we have  
<math>\int \arcsin y =y\arcsin y -\int \frac{y}{\sqrt{1-y^2}}</math>.
+
<math>\int \arcsin y dy=y\arcsin y -\int \frac{y}{\sqrt{1-y^2}}dy</math>.
  
The second integral on the right can be evaluated as <math> \int \frac{y}{\sqrt{1-y^2}}= -\sqrt{1-y^2}+C </math> for some constant <math>C </math>.
+
The second integral on the right can be evaluated as <math> \int \frac{y}{\sqrt{1-y^2}}dy= -\sqrt{1-y^2}+C </math> for some constant <math>C </math>.
  
 
Thus the answer is  <math>\color{blue} y\arcsin y+\sqrt{1-y^2}+C , \text{ where } C \text{ is a constant.}</math>
 
Thus the answer is  <math>\color{blue} y\arcsin y+\sqrt{1-y^2}+C , \text{ where } C \text{ is a constant.}</math>

Latest revision as of 22:25, 8 March 2018

We use the substitution , then

Then using integration by parts we have .

The second integral on the right can be evaluated as for some constant .

Thus the answer is