Evaluate the integral and express your answer in simplest form. int(4)/(16+25x^(2))dx Answer:

Recognize Trig Substitution: We are given the integral to evaluate: int(4)/(16+25x^(2))dx To simplify the integral, we can recognize that the denominator has the form of a^2 + u^2, where a is a constant and u is a function of x. This suggests that we can use a trigonometric substitution, specifically the tangent substitution, where x = (a/b)tan(θ) and dx = (a/b)sec^2(θ)dθ. In our case, a^2 = 16 and b^2 = 25, so a = 4 and b = 5. Therefore, we can let x = (4/5)tan(θ), which gives us dx = (4/5)sec^2(θ)dθ.Substitute x and dx: Now we substitute x and dx in the integral: int(4)/(16+25x^(2))dx = int(4)/(16+25*(4/5)^2*tan^2(θ)) * (4/5)sec^2(θ)dθ Simplify the expression inside the integral: int(4)/(16+25*(16/25)*tan^2(θ)) * (4/5)sec^2(θ)dθ = int(4)/(16+16*tan^2(θ)) * (4/5)sec^2(θ)dθ This simplifies to: int(4)/(16*(1+tan^2(θ))) * (4/5)sec^2(θ)dθ Since 1 + tan^2(θ) = sec^2(θ), we can further simplify: int(4)/(16*sec^2(θ)) * (4/5)sec^2(θ)dθ = int(4)/(4/5)dθSimplify Expression: We can now cancel out the sec^2(θ) terms and simplify the constants: int(4)/(4/5)dθ = int(5)dθ This integral is straightforward to evaluate: int(5)dθ = 5θ + CEvaluate Integral: We now need to substitute back for θ using our original substitution x = (4/5)tan(θ). To find θ, we take the arctangent of both sides: θ = arctan(5/4 * x) So our integral in terms of x is: 5θ + C = 5*arctan(5/4 * x) + C

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Evaluate the integral and express your answer in simplest form.

int(4)/(16+25x^(2))dx
Answer:

Evaluate the integral and express your answer in simplest form. $\newline$ $\int \frac{4}{16+25 x^{2}} d x$ $\newline$ Answer:

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Evaluate definite integrals using the power rule

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Q. Evaluate the integral and express your answer in simplest form.

\newline

\int \frac{4}{16+25 x^{2}} d x

\newline

Answer:

Recognize Trig Substitution: We are given the integral to evaluate: $\newline$ $\int \frac{4}{16+25x^{2}}dx$ $\newline$ To simplify the integral, we can recognize that the denominator has the form of $a^2 + u^2$ , where $a$ is a constant and $u$ is a function of $x$ . This suggests that we can use a trigonometric substitution, specifically the tangent substitution, where $x = \frac{a}{b}\tan(\theta)$ and $dx = \frac{a}{b}\sec^2(\theta)d\theta$ . In our case, $a^2 = 16$ and $b^2 = 25$ , so $a = 4$ and $a^2 + u^2$ $0$ . Therefore, we can let $a^2 + u^2$ $1$ , which gives us $a^2 + u^2$ $2$ .
Substitute $x$ and $dx$ : Now we substitute $x$ and $dx$ in the integral: $\newline$ $\int\frac{4}{16+25x^{2}}dx = \int\frac{4}{16+25\left(\frac{4}{5}\right)^2\tan^2(\theta)} \left(\frac{4}{5}\right)\sec^2(\theta)d\theta$ $\newline$ Simplify the expression inside the integral: $\newline$ $\int\frac{4}{16+25\left(\frac{16}{25}\right)\tan^2(\theta)} \left(\frac{4}{5}\right)\sec^2(\theta)d\theta = \int\frac{4}{16+16\tan^2(\theta)} \left(\frac{4}{5}\right)\sec^2(\theta)d\theta$ $\newline$ This simplifies to: $\newline$ $\int\frac{4}{16\left(1+\tan^2(\theta)\right)} \left(\frac{4}{5}\right)\sec^2(\theta)d\theta$ $\newline$ Since $1 + \tan^2(\theta) = \sec^2(\theta)$ , we can further simplify: $\newline$ $\int\frac{4}{16\sec^2(\theta)} \left(\frac{4}{5}\right)\sec^2(\theta)d\theta = \int\frac{4}{\frac{4}{5}}d\theta$
Simplify Expression: We can now cancel out the $\sec^2(\theta)$ terms and simplify the constants: $\newline$ $\int \frac{4}{\left(\frac{4}{5}\right)}d\theta = \int 5d\theta$ $\newline$ This integral is straightforward to evaluate: $\newline$ $\int 5d\theta = 5\theta + C$
Evaluate Integral: We now need to substitute back for $\theta$ using our original substitution $x = \frac{4}{5}\tan(\theta)$ . To find $\theta$ , we take the arctangent of both sides: $\newline$ $\theta = \arctan(\frac{5}{4} \times x)$ $\newline$ So our integral in terms of $x$ is: $\newline$ $5\theta + C = 5\times\arctan(\frac{5}{4} \times x) + C$