Answer to Question #122831 in Calculus for Prateeksha vats

Question

Integrate the following function w.r.t. x using substitution : 1) x^3√x^4+1 dx . lim 0 to 1. 2) tan^7x sec^2x dx lim 0 to π/4 . 3) sec^2(cosx) sinx dx

Answer 1

1. $\int_0^1x^3\sqrt{x^4+1}\,\,dx=\frac14\int_0^1\sqrt{x^4+1}\,\,d(x^4+1)=\frac{1}{6} (x^4+1)^{\frac32}\vert_{0}^{1}=\frac{2\sqrt{2}-1}{6}\approx 0.3047$
2. $\int_0^{\frac{\pi}{4}}\frac{tan^7\,x}{cos^2\,x}\,\,dx=\int_0^{\frac{\pi}{4}}{tan^7\,x}\,\,d(tan\,x)=\frac{1}{8} tan^8\,x\vert_{0}^{\frac{\pi}{4}}=\frac18=0.125$
3. $\int{\frac{sin\,x}{cos^2(cos\,x)}}\,\,dx=-\int{\frac{1}{cos^2(cos\,x)}}d(cos\,x)=-tan(cos\,x)+C,\,\, C\in{\mathbb{R}}.$

We used changes of variables (known also as integration by substitution) in integrals in all tasks(see e.g., https://en.wikipedia.org/wiki/Integration_by_substitution). The value in the first task was rounded to 4 decimal places.

Answers: 1. $\frac{2\sqrt{2}-1}{6}\approx 0.3047$ (it is rounded to 4 decimal places);

2. 0.125;

3. $=-tan(cos\,x)+C,\,\, C\in{\mathbb{R}}$.