Question

1. find the indicated antiderivatives

a) ( int 3 sqrt3{x}+2 cos x-sec ^{2} x d x )
b) ( int \frac{3}{x^{4}}+4 e^{-4 x}+\frac{1}{sqrt{x}} d x )

Explanation:

Part a)

Step1: Split the integral

$$\int 3\sqrt[3]{x}+2\cos x-\sec^{2}x\;dx=\int 3x^{\frac{1}{3}}dx+\int 2\cos x\;dx-\int\sec^{2}x\;dx$$

Step2: Integrate each term

• For $\int 3x^{\frac{1}{3}}dx$: Use the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n

eq - 1$). Here $n=\frac{1}{3}$, so $\int 3x^{\frac{1}{3}}dx=3\times\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}+C_1=\frac{9}{4}x^{\frac{4}{3}}+C_1$

• For $\int 2\cos x\;dx$: Since $\int\cos x\;dx=\sin x + C$, then $\int 2\cos x\;dx=2\sin x+C_2$
• For $\int\sec^{2}x\;dx$: We know that $\int\sec^{2}x\;dx=\tan x + C$, so $-\int\sec^{2}x\;dx=-\tan x+C_3$

Step3: Combine the results

Combining the three results: $\frac{9}{4}x^{\frac{4}{3}}+2\sin x-\tan x + C$ (where $C = C_1+C_2+C_3$)

Part b)

Step1: Split the integral

$$\int\frac{3}{x^{4}}+4e^{-4x}+\frac{1}{\sqrt{x}}\;dx=\int 3x^{-4}dx+\int 4e^{-4x}\;dx+\int x^{-\frac{1}{2}}dx$$

Step2: Integrate each term

• For $\int 3x^{-4}dx$: Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n

eq - 1$), with $n=-4$, we get $\int 3x^{-4}dx=3\times\frac{x^{-4 + 1}}{-4+1}+C_1=-x^{-3}+C_1$

• For $\int 4e^{-4x}\;dx$: Let $u=-4x$, then $du=-4dx$. So $\int 4e^{-4x}\;dx=-\int e^{u}du=-e^{-4x}+C_2$
• For $\int x^{-\frac{1}{2}}dx$: Using the power rule with $n =-\frac{1}{2}$, we have $\int x^{-\frac{1}{2}}dx=\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C_3=2x^{\frac{1}{2}}+C_3$

Step3: Combine the results

Combining the three results: $-\frac{1}{x^{3}}-e^{-4x}+2\sqrt{x}+C$ (where $C = C_1+C_2+C_3$)

Answer:

a) $\frac{9}{4}x^{\frac{4}{3}}+2\sin x-\tan x + C$

b) $-\frac{1}{x^{3}}-e^{-4x}+2\sqrt{x}+C$