Question

How do you integrate `int sqrt(1-7x^2)` using trig substitutions?

Answer 1

`int \ sqrt(1-7x^2) \ dx = 1/(2sqrt(7)) arcsin(sqrt(7)x) + 1/2 x sqrt(1-7x^2) +C`

Explanation:

Let:

`I = int \ sqrt(1-7x^2) \ dx`

We would aim to get an expression involving `a^2-x^2` prior to substitution so we can rewrite as:

`I = int \ sqrt(1-(sqrt(7)x)^2) \ dx`

If we compare to the trig identity `1-sin^2theta -= cos^2theta` then it may be worth trying the substitution:

`sin theta = sqrt(7)x => theta = arcsin(sqrt(7)x)`
and, `cos theta (d theta)/dx = sqrt(7)`

Substituting into our integral gives:

`I = int \ sqrt(1-sin^2 theta) \ (cos theta/sqrt(7)) \ d theta`
`\ \ = 1/sqrt(7) \ int \ sqrt(cos^2 theta) cos theta \ d theta`
`\ \ = 1/sqrt(7) \ int \ cos^2 theta \ d theta`

Now we use the identity:

`cos2theta -= cos^2theta - sin^2 theta`
`" " = cos^2theta - (1-cos^2) theta`
`" " = 2cos^2theta - 1`

Which gives us:

`I = 1/sqrt(7) \ int \ cos^2 theta \ d theta`
`\ \ = 1/sqrt(7) \ int \ 1/2(1+cos2theta) \ d theta`
`\ \ = 1/(2sqrt(7)) \ int \ 1+cos2theta \ d theta`
`\ \ = 1/(2sqrt(7)) { theta + 1/2sin2theta } +C`
`\ \ = 1/(2sqrt(7)) theta + 1/(4sqrt(7))sin2theta +C`
`\ \ = 1/(2sqrt(7)) theta + 1/(4sqrt(7))(2sinthetacostheta) +C`

And if we use:

`costheta = sqrt(1-sin^2theta)`

then we can restore the substitution and get:

`I = 1/(2sqrt(7)) theta + 1/2sinthetasqrt(1-sin^2theta) +C`
`\ \ = 1/(2sqrt(7)) arcsin(sqrt(7)x) + 1/(2sqrt(7))(sqrt(7)x)sqrt(1-(sqrt(7)x^2)) +C`
`\ \ = 1/(2sqrt(7)) arcsin(sqrt(7)x) + 1/2 x sqrt(1-7x^2) +C`