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Thread: Integration by substitution

  1. #1
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    Integration by substitution

    Hi,

    Is this correct? Else, may I have some assistance please?

    If $\displaystyle x = 4\cos^2 \theta + 7 \sin^2\theta$, show that $\displaystyle 7 - x = 3\cos^2\theta$, and find a similar expression for $\displaystyle x - 4$. By using substitution $\displaystyle x = 4cos^2\theta + 7\sin^2\theta$, evaluate $\displaystyle \int_4^7 \frac{1}{\sqrt{x - 4)}(7 - x)}\,dx$

    $\displaystyle x = 4\cos^2\theta + 7\sin^2\theta$

    $\displaystyle x = 4\cos^2\theta + 7(1 - cos^2\theta)$

    $\displaystyle ( x = 4\cos^2\theta + 7 - 7cos^2\theta) \times -1$

    $\displaystyle - x = - 4\cos^2\theta - 7 + 7cos^2\theta$

    $\displaystyle 7 - x = 3 \cos^2\theta$

    $\displaystyle x = 4\cos^2\theta + 7 \sin^2\theta$

    $\displaystyle x = 4(1 - \sin^2\theta) + 7\sin^2\theta$

    $\displaystyle x = 4 - 4\sin^2\theta + 7 \sin^2\theta$

    $\displaystyle x - 4 = 3 \sin^2\theta$


    $\displaystyle \int_4^7 \frac{1}{\sqrt{(x - 4)(7 - x)}}d\theta$

    $\displaystyle \int_4^7 \frac{1}{\sqrt{(3\sin^2\theta)(3\cos^2\theta)}}d\t heta$

    $\displaystyle \int_0^{90} \frac{6\cos\theta\sin\theta}{\sqrt{(3\sin^2\theta) (3\cos^2\theta)}}d\theta$

    $\displaystyle \int_0^{90} \frac{3\sin 2\theta}{\sqrt{9\cos^2\theta - 9\cos^4\theta}}$

    $\displaystyle \int_0^{90} \frac{3\sin2\theta}{3\sqrt{cos^2\theta - \cos^4\theta}}d\theta$

    $\displaystyle \int_0^{90} \frac{\sin 2\theta}{\sqrt{\cos^2\theta(1 - cos^2\theta)}}d\theta$

    $\displaystyle \int_0^{90} \frac{2\sin\theta}{cos\theta\sqrt{(1 - cos^2\theta)}}d\theta$

    $\displaystyle 2 \sin\theta \int_0^{90} \frac{1}{1 - \cos^2\theta}d\theta$

    $\displaystyle \big[2\sin\theta \cdot \sin^{-1}(\cos\theta\big]_0^{90}$

    $\displaystyle [2\cdot 1 \cdot \sin^{-1}(0)] - [0 \cdot 0]$

    $\displaystyle 0$
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  2. #2
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    Quote Originally Posted by Hellbent View Post
    Hi,

    Is this correct? Else, may I have some assistance please?

    If $\displaystyle x = 4\cos^2 \theta + 7 \sin^2\theta$, show that $\displaystyle 7 - x = 3\cos^2\theta$, and find a similar expression for $\displaystyle x - 4$. By using substitution $\displaystyle x = 4cos^2\theta + 7\sin^2\theta$, evaluate $\displaystyle \int_4^7 \frac{1}{\sqrt{x - 4)}(7 - x)}\,dx$

    $\displaystyle x = 4\cos^2\theta + 7\sin^2\theta$

    $\displaystyle x = 4\cos^2\theta + 7(1 - cos^2\theta)$

    $\displaystyle ( x = 4\cos^2\theta + 7 - 7cos^2\theta) \times -1$

    $\displaystyle - x = - 4\cos^2\theta - 7 + 7cos^2\theta$

    $\displaystyle 7 - x = 3 \cos^2\theta$

    $\displaystyle x = 4\cos^2\theta + 7 \sin^2\theta$

    $\displaystyle x = 4(1 - \sin^2\theta) + 7\sin^2\theta$

    $\displaystyle x = 4 - 4\sin^2\theta + 7 \sin^2\theta$

    $\displaystyle x - 4 = 3 \sin^2\theta$


    So far so good...


    $\displaystyle \int_4^7 \frac{1}{\sqrt{(x - 4)(7 - x)}}d\theta$

    $\displaystyle \int_4^7 \frac{1}{\sqrt{(3\sin^2\theta)(3\cos^2\theta)}}d\t heta$


    Erase this step: if you already changed variables you must already change dx by $\displaystyle d\theta$ and also the integral's limits


    $\displaystyle \int_0^{90} \frac{6\cos\theta\sin\theta}{\sqrt{(3\sin^2\theta) (3\cos^2\theta)}}d\theta$


    !!! Haven't you been told that you can not use degrees when dealing with differential/integral calculus and trigonometric

    function but you must shift to radians?! The limits must be $\displaystyle \int\limits_0^{\pi/2}$



    $\displaystyle \int_0^{90} \frac{3\sin 2\theta}{\sqrt{9\cos^2\theta - 9\cos^4\theta}}$

    $\displaystyle \int_0^{90} \frac{3\sin2\theta}{3\sqrt{cos^2\theta - \cos^4\theta}}d\theta$

    $\displaystyle \int_0^{90} \frac{\sin 2\theta}{\sqrt{\cos^2\theta(1 - cos^2\theta)}}d\theta$

    $\displaystyle \int_0^{90} \frac{2\sin\theta}{cos\theta\sqrt{(1 - cos^2\theta)}}d\theta$

    $\displaystyle 2 \sin\theta \int_0^{90} \frac{1}{1 - \cos^2\theta}d\theta$

    $\displaystyle \big[2\sin\theta \cdot \sin^{-1}(\cos\theta\big]_0^{90}$

    $\displaystyle [2\cdot 1 \cdot \sin^{-1}(0)] - [0 \cdot 0]$

    $\displaystyle 0$

    I'm afraid you messed up big time with the easiest part of the solution, after dealing nicely with the hardest part!

    You already had $\displaystyle \int\limits_0^{\pi/2} \frac{6\cos\theta\sin\theta}{\sqrt{(3\sin^2\theta) (3\cos^2\theta)}}d\theta=\int\limits_0^{\pi/2}\frac{6\cos\theta\sin\theta}{3\cos\theta\sin\the ta}d\theta=2\int\limits_0^{\pi/2}d\theta=\pi$...as simple as that!

    Tonio
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