Need help with integral substitution

[nova mauve]

what is 1/cos(x)?

[harsh glen]

$$\int \frac{tan(z+5)}{cos^2(z+5)}:dz$$

[heavy parrot]

böserübeltäter

[harsh glen]

Here are my two solution:

$$\int_{}^{}\frac{tan(z+5)}{cos^{2}(z+5)}:dz::::::u=tan(z+5)::::\frac{du}{dz}=\frac{1}{cos^2(z+5)}\Leftrightarrow dz=cos^2(z+5):du\$$

[heavy parrot]

böserübeltäter

[harsh glen]

$$\int \frac{u}{cos^2(z+5)}cos^2(z+5):du=\int u:du=\frac{1}{2}u^2 +c = \frac12tan^2(z+5)+c\$$

[heavy parrot]

böserübeltäter

[harsh glen]

$$tan(z+5)=\frac{sin(z+5)}{cos(z+5)}\Leftrightarrow tan^2(z+5)= \frac{sin^2(z+5)}{cos^2(z+5)}\Leftrightarrow \$$

[heavy parrot]

böserübeltäter

[harsh glen]

$$\frac12tan^2(z+5)+c=\frac12tan^2(z+5)+c
$$

[heavy parrot]

böserübeltäter

[harsh glen]

And my second solution:

$$\int_{}^{}\frac{tan(z+5)}{cos^{2}(z+5)}:dz=\int \frac{sin(z+5)}{cos(z+5)}\frac{1}{cos^2(z+5)}:dz=\int \frac{sin(z+5)}{cos^3(z+5)}$$

[heavy parrot]

böserübeltäter

[harsh glen]

$$u=cos(z+5)::::\frac{du}{dz}=-sin(z+5)\Leftrightarrow dz=-\frac{du}{sin(z+5)}$$

[heavy parrot]

böserübeltäter

[harsh glen]

$$=-\int \frac{sin(z+5)}{u^3}\frac{du}{sin(z+5)}=-\int \frac{1}{u^3}:du=-\int u^{-3}:du=\frac12u^{-2}+c=\frac12cos^{-2}(z+5)+c=\frac12\frac{1}{cos^2(z+5)}+c$$

[heavy parrot]

böserübeltäter

[nova mauve]

uh, just substitute tan(x)=u

[harsh glen]

$$tan(z+5)=\frac{sin(z+5)}{cos(z+5)}\Leftrightarrow tan^2(z+5)= \frac{sin^2(z+5)}{cos^2(z+5)}\Leftrightarrow cos^2(z+5)=\frac{sin^2(z+5)}{tan^2(z+5)}$$

[heavy parrot]

böserübeltäter

[harsh glen]

$$\frac12\frac{1}{cos^2(z+5)}+c=\frac12\frac{tan(z+5)}{sin^2(z+5)}+c$$

[heavy parrot]

böserübeltäter