Compute [math: ∫π22tan−1⁡3−3−7cot⁡x1+sin⁡x dx]\displaystyle \int _{\frac {\pi }{2}}^{2 an ^{-1}{3}}{\frac {-3-7\cot {x}}{1+\sin {x}} dx}. (Correct the answer to 2 decimal places.)

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Compute [math: ∫π22tan−1⁡3−3−7cot⁡x1+sin⁡x dx]\displaystyle \int _{\frac {\pi }{2}}^{2	an ^{-1}{3}}{\frac {-3-7\cot {x}}{1+\sin {x}} dx}. (Correct the answer to 2 decimal places.)

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The problem presents a definite integral with a single expression to evaluate: ∫ from π/2 to 2 arctan(3) of (-3 - 7 cot x) / (1 + sin x) dx. Since there are no multiple-choice options provided, I will outline a clear path to evaluate the integral and then discuss how one would obtain a numerical result to two decimal places.

First, transform the trigonometric expressions into a rational form that is easier to integrate. A standard approach is to use the Weierstrass substitution t = tan(x/2). Under this substitution:
- sin x = 2t/(1+t^2)
- cos x = (1 - t^2)/(1+t^2)
- dx = 2 dt/(1+t^2)
- cot x = cos x / sin x = [(1 - t^2)/(1+t^2)] / [2t/(1+t^2)] = (1 - t^2)/(2t)
- When x = π/2, t = tan(π/4) = 1; when x = 2 arctan(3), t = tan((2 arctan 3)/2) = tan(arctan 3) = 3.

With these, the integrand becomes a rational function in t. Specifically, (-3 - 7 cot x) / (1 + sin x) turns into a rational expression in t, and dx contributes another factor of 2 dt/(1+t^2). After substitution, the integral becomes an integral in t from 1 to 3 of a rational function of t. Carrying out the algebra gives a simplified integrand of the form
∫ from t=1 to t=3 of [(-6) - (7/t) + 7t] / (t+1)^2 dt.

Second, integrate the resulting rational function. The decomposition can be done by standard partial fraction techniques or by recognizing the integrand as a derivative of a simpler quotient. One convenient path is to rewrite the integrand as a sum of simpler fractions whose antiderivatives are straightforward, or to perform polynomial division followed by partial fractions. In practice, carrying out the algebra yields a closed-form expression evaluated between t = 1 and t = 3.

Third, evaluate the definite integral numerically to two decimal places. After performing the integration (whether by exact antiderivative and then substitution back to x, or by a precise numerical method on the t-integral), you obtain a numerical value close to 0.51. Rounding to two decimals gives 0.51.

Notes on potential pitfalls and common misconceptions:
- If one attempts to differentiate or integrate in x directly without the tangent half-angle substitution, the algebra can become unwieldy due to cot x and sin x in the same denominator.
- It is important to correctly transform the limits under t = tan(x/2); mixing up t-values will yield an entirely different result.
- When performing the algebra, keep track of factors of (1+t^2) that cancel between the numerator, denominator, and dx factor; a small algebraic slip can lead to a wrong sign or coefficient.

In summary, by converting to t = tan(x/2), simplifying the resulting rational integrand, integrating, and evaluating between t = 1 and t = 3, the value of the integral is obtained as approximately 0.51 to two decimal places.

_MATH1013_1ABCD_2025 Subsection 5.2 (closed on 15 Nov)

Compute [math: ∫π22tan−1⁡3−3−7cot⁡x1+sin⁡x dx]\displaystyle \int _{\frac {\pi }{2}}^{2\tan ^{-1}{3}}{\frac {-3-7\cot {x}}{1+\sin {x}} dx}. (Correct the answer to 2 decimal places.)

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