# How to solve this integral without reduction formula?

Integrate $$\int(7t^2-3)^\frac{5}{2}dt$$

My attempt,

Let $t=\sqrt{\frac{3}{7}}\sec \theta$

$dt=\sqrt{\frac{3}{7}}\sec \theta \tan \theta d\theta$

$$=\int(7\cdot \frac{3}{7}\sec^2 \theta -3)^\frac{5}{2} \cdot \sqrt{\frac{3}{7}}\sec \theta \tan \theta d\theta$$

$$=\int(3\tan^2\theta)^{\frac{5}{2}} \cdot \sqrt{\frac{3}{7}}\sec \theta \tan \theta d\theta$$

$$\int\frac{27\sqrt{7}}{7}\sec \theta \tan^6 \theta d\theta$$

$$=\int\frac{27\sqrt{7}}{7}\sec \theta (\sec^2 \theta -1)^3d\theta$$

$$\frac{27\sqrt{7}}{7}\int sec^7 \theta-3\sec^5 \theta+3\sec^3 \theta -\sec \theta d\theta$$

I know I could proceed my attempt with reduction formula for $\sec^n \theta$, but reduction formula is not included in my syllabus. How could this question be solved differently? Thanks in advance.

• Integrate by parts – lab bhattacharjee Oct 17 '17 at 14:00
• @labbhattacharjee How? – Mathxx Oct 17 '17 at 14:07

Improve your solution from $$\int\frac{27\sqrt{7}}{7}\sec \theta \tan^6 \theta d\theta=\int\frac{27\sqrt{7}}{7}\dfrac{\sin^6\theta}{(1-\sin^2\theta)^4} \cos\theta\,d\theta$$ and let $u=\sin\theta$.