He plugged in back in as 1-sin^2 but shouldn't it be 1-(1/sqrt(7))sin^2)?
Why is this trigonometric substitution plugged in without the constant?
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$\begingroup$
calculus
integration
trigonometric-integrals
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3No, he plugged it into $1-7w^2$ not $1-w^2$. – 2017-02-24
1 Answers
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Well, you are substituting $w=\frac{1}{\sqrt{7}}\sin{\theta}$, therefore: $$\sqrt{1-7w^2}=\sqrt{1-7\cdot \left(\frac{1}{\sqrt{7}}\sin{\theta}\right)^2}=\sqrt{1-7\cdot \left(\frac{1}{7}\sin^2{\theta}\right)}=\sqrt{1-\sin^2{\theta}}=\cos{\theta}$$
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0I see my mistake now, I wasn't squaring the whole value, thanks. – 2017-02-24
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0@cout You're welcome. Yes, sometimes it is easy to forget about that. – 2017-02-24