Question

Evaluate ( I=int frac{3 sin heta}{2 sqrt{4-cos heta}} d heta ). Evaluate ( I=int frac{1-2 x}{sqrt{1+2 x-2 x^{2}}} d x ). Evaluate ( I=int frac{9}{(1-3 x)^{4}} d x ). Evaluate ( I=int sqrt[3]{x^{3}+1} x^{5} d x )

          Evaluate ( I=int frac{3 sin 	heta}{2 sqrt{4-cos 	heta}} d 	heta ).
Evaluate ( I=int frac{1-2 x}{sqrt{1+2 x-2 x^{2}}} d x ).
Evaluate ( I=int frac{9}{(1-3 x)^{4}} d x ).
Evaluate ( I=int sqrt[3]{x^{3}+1} x^{5} d x )

$Evaluate ( I=int frac3 sin heta2 sqrt4-cos heta d heta ). Evaluate ( I=int frac1-2 xsqrt1+2 x-2 x^2 d x ). Evaluate ( I=int frac9(1-3 x)^4 d x ). Evaluate ( I=int sqrt[3]x^3+1 x^5 d x )$

Added by Bafana B.

Calculus: Early Transcendentals

William Briggs, Lyle Cochran, Bernard… 3rd Edition

Chapter 8

Instant Answer

Solved by Expert Keondre Parker

08/17/2023

Step 1

For the first integral, we can use the substitution method. Let $u = 2 - \cos\theta$, then $du = \sin\theta d\theta$. The integral becomes: \[I = \frac{3}{2} \int \frac{du}{\sqrt{u}} = \frac{3}{2} \cdot 2\sqrt{u} + C = 3\sqrt{2 - \cos\theta} + C\] Show more…

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