Question

Use trigonometric substitution to evaluate the integral 13 + 12x − x2 dx . First, write the expression under the radical in an appropriate form so that a trigonometric substitution can be performed. 13 + 12x − x2

Answer 1

I don't see a square root sign anywhere, so I'll assume the integral is

$\displaystyle\int\sqrt{13+12x-x^2}\,\mathrm dx$

First complete the square:

$13+12x-x^2=49-(6-x)^2=7^2-(6-x)^2$

Now in the integral, substitute

$6-x=7\sin t\implies\mathrm dx=-7\cos t\,\mathrm dt$

so that

$t=\sin^{-1}\left(\dfrac{6-x}7\right)$

Under this change of variables, we have

$7^2-(6-x)^2=7^2-7^2\sin^2t=7^2(1-\sin^2t)=7^2\cos^2t$

so that

$\displaystyle\int\sqrt{13+12x-x^2}\,\mathrm dx=-7\int\sqrt{7^2\cos^2t}\,\cos t\,\mathrm dt=-49\int|\cos t|\cos t\,\mathrm dt$

Under the right conditions, namely that cos(t) > 0, we can further reduce the integrand to

$|\cos t|\cos t=\cos^2t=\dfrac{1+\cos(2t)}2$

$\displaystyle-49\int|\cos t|\cos t\,\mathrm dt=-\frac{49}2\int(1+\cos(2t))\,\mathrm dt=-\frac{49}2\left(t+\frac12\sin(2t)\right)+C$

Expand the sine term as

$\dfrac12\sin(2t)}=\sin t\cos t$

Then

$t=\sin^{-1}\left(\dfrac{6-x}7\right)\implies \sin t=\dfrac{6-x}7$

$t=\sin^{-1}\left(\dfrac{6-x}7\right)\implies \cos t=\sqrt{7^2-(6-x)^2}=\sqrt{13+12x-x^2}$

So the integral is

$\displaystyle-\frac{49}2\left(\sin^{-1}\left(\dfrac{6-x}7\right)+\dfrac{6-x}7\sqrt{13+12x-x^2}\right)+C$