Question

Integrate w.r.t to x using substitution
(A) ∫ x^3√x^4+1 dx from 1 to 0
(B) tan^7x sec^2x dx from pi by 4 to 0
(C) sec^2 (cosx) sinx dx

Answer 1

$A.) \int_1^0 x^3 \sqrt{x^4 + 1} \, dx$

Let us assume that:

$x^4 + 1 = u \\ \Rightarrow 4x^3 \, dx = du$

$\Rightarrow x^3 \, dx =\frac{du}{4}$

When:

$x = 1$

Then:

$u = 1^4 + 1 = 2$

Again, when:

$x = 0$

Then:

$u = 0^4 + 1 = 1$

Substituting $u = x^4 + 1$ into the integration and integrating with respect to $du$ , we have:

$= \frac{1}{4} \int_2^1 \sqrt{u} \, du$

$= \frac{1}{4} \int_2^1 u^{\frac{1}{2}} \, du$

$= \frac{1}{4} \left[ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right]_2^1$ $\hspace{1 cm} \left[ \because \int x^n \, dx = \frac{x^{n+1}}{n+1} \right]$

$= \frac{1}{6} \left[ u^{\frac{3}{2}} \right]_2^1$

$= \frac{1}{6} \left[ 1^{\frac{3}{2}} - 2^{\frac{3}{2}} \right] \\ = \frac{1 - 2\sqrt{2}}{6}$

$\\$

B.) $\int_{\frac{\pi}{4}}^0 \tan^7 (x) \sec^2 (x) \, dx$

Let us assume that:

$tan (x) = u \\ \Rightarrow sec^2 (x) \, dx = du$

When:

$x = \frac{\pi}{4}$

Then:

$u = \tan (\frac{\pi}{4}) = 1$

Again, when:

$x = 0$

Then:

$u = \tan (0) = 0$

Substituting $u = \tan (x)$ into the integration and integrating with respect to $du$ , we have:

$= \int_1^0 u^7 \, du \\ = \frac{1}{8} \left[ u^8 \right]_1^0$ $\hspace{1 cm} \left[ \because \int x^n \, dx = \frac{x^{n+1}}{n+1} \right]$

$= - \frac{1}{8}$

$\\$

C.) $\int \sec^2 (\cos(x)) \sin (x) \, dx$

Let us assume that:

$\cos(x) = u \\ \Rightarrow - \sin (x) \, dx = du$

Substituting $u = \cos (x)$ into the integration and integrating with respect to $du$ , we have:

$= - \int sec^2 (u) \, du \\ = - tan (u) + C \hspace{1 cm}$ $\left[ \because \int \sec^2 (x) \, dx = \tan (x) + C \right]$

Where C is a constant of integration.

Undo substitution and we have:

$= - \tan (\cos(x)) + C$ $\hspace{1 cm} \left[ \because u = \cos(x) \right]$