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$f(x) = \int_{1}^{u} \cos^2 t \, dt$ , where $u=\ln(x^2 +x-1)$

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Transformations of functions

Full solution

Q.

f(x) = \int_{1}^{u} \cos^2 t \, dt

, where

u=\ln(x^2 +x-1)

Apply Power-Reduction Formula: To solve the integral of $\cos^2(t)$ , we can use a power-reduction formula from trigonometry, which states that $\cos^2(t)$ can be expressed as $(1 + \cos(2t))/2$ . This will simplify the integration process.
Rewrite Integral Using Formula: Now, we rewrite the integral using the power-reduction formula: $\newline$ $f(x) = \int_{1}^{\ln(x^2 + x - 1)} \frac{1 + \cos(2t)}{2} \, dt$ .
Split Integral into Two: Next, we split the integral into two separate integrals: $f(x) = \frac{1}{2}\int_{1}^{\ln(x^2 + x - 1)} 1 \, dt + \frac{1}{2}\int_{1}^{\ln(x^2 + x - 1)} \cos(2t) \, dt$ .
Integrate Each Part Separately: We can now integrate each part separately. The integral of $1$ with respect to $t$ is simply $t$ , and the integral of $\cos(2t)$ with respect to $t$ is $(1/2)\sin(2t)$ because the derivative of $\sin(2t)$ is $2\cos(2t)$ , so we need to multiply by $1/2$ to compensate for the derivative of the inside function. $\newline$ $f(x) = (1/2)[t]$ from $1$ to $t$ $1$ + $t$ $2$ from $1$ to $t$ $1$ .
Evaluate Integrals at Limits: Now we evaluate the integrals at the upper and lower limits: $f(x) = \frac{1}{2}[\ln(x^2 + x - 1) - 1] + \frac{1}{4}[\sin(2\ln(x^2 + x - 1)) - \sin(2)]$ .
Simplify the Expression: Finally, we simplify the expression to get the final answer: $f(x) = \frac{1}{2}\ln(x^2 + x - 1) - \frac{1}{2} + \frac{1}{4}\sin(2\ln(x^2 + x - 1)) - \frac{1}{4}\sin(2)$ .

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