Problem

(a) Prove the reduction formula $$ \int \cos^n x…

08:10

Problem 47 Hard Difficulty

(a) Use the reduction formula in Example 6 to show that
$$ \int \sin^2 x dx = \frac{x}{2} - \frac{\sin 2x}{4} + C $$
(b) Use part (a) and the reduction formula to evaluate $ \displaystyle \int \sin^4 x dx $.

Answer

a) $\int \sin ^{2} x d x=\frac{x}{2}-\frac{\sin (2 x)}{4}+C$
b) $\int \sin ^{4} x d x=\frac{3 x}{8}-\frac{1}{4} \cos x \sin ^{3} x-\frac{3 \sin (2 x)}{16}+C$

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Video Transcript

Hello. Welcome to this lesson. This lesson. We'll use the reduction, the reduction formula to show that going on. Yeah. The integral sign too. A science squared X The X is equal to X on two minus sign to X on four. Blast. See? Yeah. Yeah. Okay, let me add a sweet would Probably the sea comes after you have integrated this. Okay, so we use this to show that wow. Mhm. So let's begin. Yeah. Yeah. What we do is that we would identify the end, then we plug it into this. Yeah, so here the n is too. So it means that the integral signs quad deitz sine squared. X DX is equal to 1/2. We are substituting into this. Go to cross X sign to the power. Um, and it's too so and minus one x than last two minus one on two than sign to minus two x dx. Oh, good. So here it becomes half cost X side X, then plus one on two. Oh, whoa. Because the whole thing becomes general and the whole of the sign is out of that. So we have half cross x sine x glass of X squared. Yeah. Yeah. Okay. This is left with DX alone. So that is the whole thing here is gone. So I just left it. Yeah. So we comes one on one on two x about from double angles. We have signed to E s. Yeah. Cost a sign. A than last sign. A course. So that gives to sign a, uh, of course, A So here we can express this. Okay. Mhm. Yeah. Yeah. So the whole thing here, it's what we have. Okay. So, divided by divided by two becomes Yeah. Oh, yeah. Okay. And these councils that so the whole of this is half of sign e costly. Okay, then we can drive that, uh, the anti ground science squared. X DX is equal to half, and the whole of this is signed two X. Okay. What? Two, then last one on two X. So this becomes pull open. One on four sides to x last one on two X as required. Now, looking at the keypad, we've been asked to find the integral sign. Four x e x. So this is simply negative. One. On what? Four. Because you've given and us for okay. Then coughs ex signed four The next one X then the last four, minus one on four, then into girls sign mm four minus two x the X we just by using the reduction formula. Okay, but how so This becomes negative One on four course back sign. Yeah. All right. Mhm. Mhm. What? Yeah, well, okay. That was negative. Yeah, Yeah, yeah, yeah, yeah, yeah, right. Yeah. Okay, so here. Wow. Yeah, yeah, yeah. Okay. Okay. Yeah. Oh, yeah. Mm hmm. Uh huh. Uh huh. Sorry. Two x dx by the whole of the integral side squared x d excess where we are found here So we can comfortably put it there. So negative on on four cause X sign three x last three on four. Then we put the whole of negative one on four sides to backs last half eggs one day, then plus see the constant of integration cause back sign three x. Okay, so this becomes negative. Three on 16. Sign two X, then glass three on eight. Ex blast. See? Mhm. So we can bring their three on eight first. Have negative one on four, cause X sign. Okay, three x then minus three on 16. Yeah, signed two X and glassy. Okay, So this is the finance. Yeah, for the interior. Signed four x x x Signed to the part four ex. Yes. So thanks for time. There's the end of the lesson.

Willis J.

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Section 1

Integration by Parts

Problem

(a) Prove the reduction formula $$ \int \cos^n x…

Problem 47 Hard Difficulty

(a) Use the reduction formula in Example 6 to show that
$$ \int \sin^2 x dx = \frac{x}{2} - \frac{\sin 2x}{4} + C $$
(b) Use part (a) and the reduction formula to evaluate $ \displaystyle \int \sin^4 x dx $.

Answer

a) $\int \sin ^{2} x d x=\frac{x}{2}-\frac{\sin (2 x)}{4}+C$
b) $\int \sin ^{4} x d x=\frac{3 x}{8}-\frac{1}{4} \cos x \sin ^{3} x-\frac{3 \sin (2 x)}{16}+C$

More Answers

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Calculus: Early Transcendentals

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Discussion

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Video Transcript

Calculus: Early Transcendentals

Anna Marie V.

Kayleah T.

Kristen K.

Michael J.

Integration Review - Intro

Definite vs Indefinite

a) $\int \sin ^{2} x d x=\frac{x}{2}-\frac{\sin (2 x)}{4}+C$b) $\int \sin ^{4} x d x=\frac{3 x}{8}-\frac{1}{4} \cos x \sin ^{3} x-\frac{3 \sin (2 x)}{16}+C$

Calculus: Early Transcendentals

Anna Marie V.

Kayleah T.

Kristen K.

Michael J.

Integration Review - Intro

Definite vs Indefinite

Anna Marie V.

Kayleah T.

Kristen K.

Michael J.

a) $\int \sin ^{2} x d x=\frac{x}{2}-\frac{\sin (2 x)}{4}+C$
b) $\int \sin ^{4} x d x=\frac{3 x}{8}-\frac{1}{4} \cos x \sin ^{3} x-\frac{3 \sin (2 x)}{16}+C$