Answer:
b) you need one set of | | for the inside of the logarithm only.
c) pi^2 /56
Step-by-step explanation:
For b, you just need | | instead of ( ) for the inside of the natural logarithm part... .
For c...
Integral( arcsin(7/6x)/sqrt(36-49x^2),x=0..6/7).
Lets go ahead and fix up the the denominator so the multipler on x is the same as it is in numerator.
Factor our sqrt(36) on bottom. Note sqrt(36)=6.
1/6Integral(arcsin(7/6x)/sqrt(1-(7x/6)^2),x=0..6/7)
Let 7/6x=sin(v). Then 7/6 dx=cos(v) dv.
If x=0 then we solve 7/6(0)=sin(v) giving us v=0. Note sin(0)=0.
If x=6/7 then we solve 7/6(6/7)=sin(v) giving us v=pi/2. Note sin(pi/2)=1...and I chose this particular value for v so the sine curve didn't have a chance to make a new or start a new cycle.
1/6×6/7Integral(v/sqrt(1-(sinv)^2) cos(v)),v=0..pi/2)
Next step invokes the Pythagorean Identity sin^2(v)+cos^2(v)=1:
1/7Integral(v/cos(v) cos(v), v=0..pi/2)
1/7Integral(v, v=0..pi/2)
1/7(v^2/2, v=0..pi/2)
1/7((pi/2)^2/2-(0)^2/2)
1/7(pi^2/4/2)
1/7(pi^2/8)
pi^2/56
