Question

Evaluate the integrals in Exercises 77–90.79. ∫(from -1 to 0)6dt/√(3-2t-t²)

Accepted Answer

First, rewrite the integral to clearly identify the integrand and the limits: $$\int_{-1}^{0} \frac{6}{\sqrt{3 - 2t - t$$^{2}$$}} \, dt. $$Complete the square inside the square root in the denominator to simplify the expression. Start with the quadratic expression: $$3 - 2t - t$$^{2}$$. Rewrite it as -(t$$^{2}$$ + 2t - 3)$$, then complete the square for $$t^{2} + 2t - 3$. Express t$$^{2}$$ + 2t - 3 in $$the form $$(t + a$$)^{2}$$ + b by $$finding $$a$$ and $$b. $$Recall that $$t^{2} + 2t = (t$$ + $$1)^{2} - 1$, so t$$^{2}$$ + 2t - 3 = (t + 1$$)^{2}$$ - 4. $$Substitute back into the integral, so the denominator becomes $$\sqrt{-( (t + 1$$)^{2}$$ - 4 )} = \sqrt{4 - (t + 1$$)^{2}$$}. $$This suggests a trigonometric substitution of the form $$t + 1 = 2 \sin 	heta. $$Change the variable of integration using the substitution $$t + 1 = 2 \sin 	heta$$, find $$dt in $$terms of $$d	heta$$, adjust the limits accordingly, and rewrite the integral in terms of $$	heta. $$Then, simplify and integrate using the standard integral formula for $$\int \frac{d	heta}{\sqrt{1 - \sin$$^{2}$$ 	heta}} or $$equivalent.

Evaluate the integrals in Exercises 77–90.
79. ∫(from -1 to 0)6dt/√(3-2t-t²)

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Key Concepts

Definite Integral

Integration of Rational Functions Involving Square Roots

Completing the Square