$$\int x(1+x)^{10} dx$$
Let \( u = 1+x \), so \( \frac{du}{dx} = 1 \)
$$\frac{du}{dx} = 1$$
$$dx = \frac{du}{1} = du$$
As \( u = 1+x \), \( x = u-1 \)
Rewrite as
$$\int (u-1)(u^{10}) du$$
$$= \int u^{11} - u^{10} du$$
$$= \frac{1}{12}u^{12} - \frac{1}{11}u^{11}$$
Substitute \( u = x+1 \) back into equation
$$\frac{1}{12}u^{12} - \frac{1}{11}u^{11}$$
$$= \frac{1}{12}(x+1)^{12} - \frac{1}{11}(x+1)^{11}$$
Integrating tan x using substitution.
Let's apply what we have learnt to integrating tanx
We know that \( \tan x = \frac{\sin x}{\cos x} \)
$$\int \tan x = \int \frac{\sin x}{\cos x} \, dx$$
Let
$$u = \cos x$$
Then
$$\frac{du}{dx} = -\sin x$$
$$dx = -\frac{1}{\sin x} \, du$$
$$\int \frac{\sin x}{u} \left( -\frac{1}{\sin x} \right) \, du = \int -\frac{1}{u} \, du = -\ln |u|$$
$$Substitute \( u = \cos x \)$$
$$-\ln |u| = -\ln |\cos x|$$CONTENT$nbsp;