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I had a test where I was asked this question. I integrated the function by taking $\tan x=u$ and then $\sec^2x \, dx=du$. I took $\sec(x) =\sqrt{1+u^2}$ and then using the basic formulas calculated the correct answer but my teacher didn't give me marks for my solution. Please explain me why. He gave me a zero.I can't post pictures otherwise I would have shown you

I got the answer $(\tan x\sec x-\log(\tan x+\sec x)/2$

Here the picture of my solution https://drive.google.com/file/d/1E9e1JoX_8C28h7k8uW6bBpoYZFy1LuCM/view?usp=drivesdk

Chanda Bagri
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  • Can you please write the exact steps you did in your question? This will help me understand where you could improve. Also, what do you mean by 'my teacher didn't give me marks'? Did he not give you any marks? – Toby Mak Nov 18 '17 at 10:09
  • Correct answer and no marks ? How come ? –  Nov 18 '17 at 10:29
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    I think you should ask your teacher. – user222031 Nov 18 '17 at 10:31
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    @ChandaBagri I understand that you can't take pictures during an exam, but can you please write down what you can remember? Show some effort, otherwise others won't show effort as well by writing an answer. – Toby Mak Nov 18 '17 at 10:35
  • I have my exam paper he gave me the paper back but didn't told me my mistake. – Chanda Bagri Nov 18 '17 at 10:46
  • @ChandaBagri : A grammatical point: The negation of "He gave" is "He didn't give", not "He didn't gave", and the negation of "He told" is "He didn't tell", not "He didn't told". – Michael Hardy Nov 19 '17 at 15:33

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\begin{align} \int\sec^3x\,dx &= \int\dfrac{1}{\cos^3x}\,dx \\ &= \int\dfrac{\cos x}{(1-\sin^2x)^2}\,dx \hspace{0.5cm};\hspace{0.5cm} \sin x=u\\ &= \int\dfrac{1}{(1-u^2)^2}\,dx \\ &= \dfrac{u}{2(1-u^2)}+\dfrac{1}{4}\ln\dfrac{1+u}{1-u}+C\\ &= \dfrac{u}{2(1-u^2)}+\dfrac{1}{4}\ln\dfrac{1+u}{1-u}+C\\ &= \dfrac{\sin x}{2\cos^2x}+\dfrac{1}{4}\ln\dfrac{1+\sin x}{1-\sin x}+C \end{align}

Nosrati
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Here's one way: \begin{align} \int \sec^3 x \, dx = {} & \int (\sec x) \Big( \sec^2 x \, dx\Big) = \overbrace{\int u\, dv = uv - \int v \, du}^{\text{integration by parts}} \\[10pt] = {} & \sec x \tan x - \int (\tan x) \big( \sec x\tan x\, dx\big) \\[10pt] = {} & \sec x \tan x - \int(\sec x)(\tan^2 x) \, dx \\[10pt] = {} & \sec x \tan x - \int(\sec x) (\sec^2 x - 1) \,dx \\[10pt] = {} & \sec x\tan x + \int \sec x\,dx - \int\sec^3 x\,dx. \\[10pt] & \text{Then adding $\int\sec^3 x\,dx$ to both sides, we get} \\[10pt] 2\int\sec^3 x\, dx & = \sec x \tan x + \int \sec x\, dx. \\[10pt] & \text{Hence} \\[10pt] \int \sec^3 x\, dx & = \frac 1 2 \sec x \tan x + \frac 1 2 \int \sec x\, dx = \cdots \end{align}

Michael Hardy
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