# How do you integrate int sqrt(7x^2-3)^(3/2) using trig substitutions?

Dec 4, 2017

$\int {\left(7 {x}^{2} - 3\right)}^{\frac{3}{2}} \cdot \mathrm{dx}$

=$\frac{7}{4} {x}^{3} \cdot \sqrt{7 {x}^{2} - 3} - \frac{15}{8} x \sqrt{7 {x}^{2} - 3} + \frac{27 \sqrt{7}}{56} \cdot \ln \left(x \sqrt{7} + \sqrt{7 {x}^{2} - 3}\right) + C$

#### Explanation:

$\int {\left(7 {x}^{2} - 3\right)}^{\frac{3}{2}} \cdot \mathrm{dx}$

After using $x = \sqrt{\frac{3}{7}} \cdot \sec u$ and $\mathrm{dx} = \sqrt{\frac{3}{7}} \cdot \sec u \cdot \tan u \cdot \mathrm{du}$ transforms, this integral became,

$\int {\left[7 \cdot \frac{3}{7} {\left(\sec u\right)}^{2} - 3\right]}^{\frac{3}{2}} \cdot \sqrt{\frac{3}{7}} \cdot \sec u \cdot \tan u \cdot \mathrm{du}$

=$\int {\left[3 {\left(\sec u\right)}^{2} - 3\right]}^{\frac{3}{2}} \cdot \sqrt{\frac{3}{7}} \cdot \sec u \cdot \tan u \cdot \mathrm{du}$

=$\int {\left[3 {\left(\tan u\right)}^{2}\right]}^{\frac{3}{2}} \cdot \sqrt{\frac{3}{7}} \cdot \sec u \cdot \tan u \cdot \mathrm{du}$

=int [3sqrt3(tanu)^3*sqrt(3/7)*secu*tanu*du

=int [9/sqrt7*(tanu)^4*secu*du

Now, I solved $\int {\left(\tan u\right)}^{4} \cdot \sec u \cdot \mathrm{du}$ integral,

$\int {\left(\tan u\right)}^{4} \cdot \sec u \cdot \mathrm{du}$

=$\int {\left[{\left(\sec u\right)}^{2} - 1\right]}^{2} \cdot \sec u \cdot \mathrm{du}$

=$\int {\left(\sec u\right)}^{5} \cdot \mathrm{du} - 2 \int {\left(\sec u\right)}^{3} \cdot \mathrm{du} + \int \sec u \cdot \mathrm{du}$

After using reduction formula for $\int {\left(\sec u\right)}^{n} \cdot \mathrm{du}$,

$\int {\left(\sec u\right)}^{n} \cdot \mathrm{du} = \frac{1}{n - 1} \cdot {\left(\sec u\right)}^{n - 2} \cdot \tan u + \frac{n - 2}{n - 1} \cdot \int {\left(\sec u\right)}^{n - 2} \cdot \mathrm{du}$

$\int {\left(\tan u\right)}^{4} \cdot \sec u \cdot \mathrm{du}$

=$\frac{1}{4} \cdot {\left(\sec u\right)}^{3} \cdot \tan u + \frac{3}{4} \int {\left(\sec u\right)}^{3} \cdot \mathrm{du} - 2 \int {\left(\sec u\right)}^{3} \cdot \mathrm{du} + \int \sec u \cdot \mathrm{du}$

=$\frac{1}{4} \cdot {\left(\sec u\right)}^{3} \cdot \tan u - \frac{5}{4} \int {\left(\sec u\right)}^{3} \cdot \mathrm{du} + \int \sec u \cdot \mathrm{du}$

=$\frac{1}{4} \cdot {\left(\sec u\right)}^{3} \cdot \tan u - \frac{5}{8} \cdot \sec u \cdot \tan u - \frac{5}{8} \int \sec u \cdot \mathrm{du} + \int \sec u \cdot \mathrm{du}$

=$\frac{1}{4} \cdot {\left(\sec u\right)}^{3} \cdot \tan u - \frac{5}{8} \cdot \sec u \cdot \tan u + \frac{3}{8} \int \sec u \cdot \mathrm{du}$

=$\frac{1}{4} \cdot {\left(\sec u\right)}^{3} \cdot \tan u - \frac{5}{8} \cdot \sec u \cdot \tan u + \frac{3}{8} \ln \left(\sec u + \tan u\right) + \frac{\sqrt{7}}{9} \cdot C 1$

Hence,

=$\int \frac{9}{\sqrt{7}} \cdot {\left(\tan u\right)}^{4} \cdot \sec u \cdot \mathrm{du}$

=$\frac{9}{4 \sqrt{7}} \cdot {\left(\sec u\right)}^{3} \cdot \tan u$-$\frac{45}{8 \sqrt{7}} \cdot \sec u \cdot \tan u$+$\frac{27}{8 \sqrt{7}} \cdot \ln \left(\sec u + \tan u\right) + C 1$

After using $x = \sqrt{\frac{3}{7}} \cdot \sec u$, $\sec u = \sqrt{\frac{7}{3}} \cdot x$ and $\tan u = \frac{\sqrt{7 {x}^{2} - 3}}{\sqrt{3}}$ inverse transformations, I found

$\int {\left(7 {x}^{2} - 3\right)}^{\frac{3}{2}} \cdot \mathrm{dx}$

=$\frac{9}{4 \sqrt{7}} \cdot 7 \frac{\sqrt{7}}{9} \cdot {x}^{3} \cdot \sqrt{7 {x}^{2} - 3}$-45/(8sqrt7)*sqrt7/3*xsqrt(7x^2-3+$\frac{27}{8 \sqrt{7}} \cdot \ln \left(\sqrt{\frac{7}{3}} \cdot x + \frac{\sqrt{7 {x}^{2} - 3}}{\sqrt{3}}\right) + C 1$

=$\frac{7}{4} {x}^{3} \cdot \sqrt{7 {x}^{2} - 3}$-15/8xsqrt(7x^2-3+$\frac{27 \sqrt{7}}{56} \cdot \ln \left(x \sqrt{7} + \sqrt{7 {x}^{2} - 3}\right) + C$

Note: $C = C 1 - \frac{27 \sqrt{7}}{112} \cdot L n 3$