1second one is even easier jst take x3 as z now d(x3)/dx =3x2 and thus apply the formulae
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∫dx 1sinx + cosx
∫x2√1+x3
∫sec7x-7sin2x
∫(1+cot(x+α)cot(x-α)) dx
∫log(cosx)cos2x
ex-1(x2-5x+4)
please try them!
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14 Answers
1for first one its simple convert sinx + cos x to √2sin(pi/4+x)
then let pi/4+x=z and continue..........
11Thanks for the clue...but can u hlp me with the derivation of the identity,,??
1the second one u can evn do it by takin the whole Denominator as some t and thn apply the formulae....but the options arent matching! :
1dude u cn take the whole denmtor as a t2 it wil smplify your wrk
and for 5th sum i think u mst cnvrt 1/cos2x as sec2 and proceed by ∫u.vdx
1708\hspace{-16}(4)\;\; \int \left\{1+\cot (x+\alpha).\cot(x-\alpha)\right\}dx\\\\ $Now Using Trigonometric Identity::\\\\ $\cot\left\{(x+\alpha)-(x-\alpha)\right\}=\frac{\cot (x+\alpha).\cot(x-\alpha)+1}{\cot (x-\alpha)-\cot (x+\alpha)}$\\\\\\ So $\boxed{1+\cot (x+\alpha).\cot(x-\alpha)=\cot(2\alpha).\left\{\cot (x-\alpha)-\cot (x+\alpha)\right\}}$\\\\\\ So Integral Convert into $\cot(2\alpha).\int \left\{\cot (x-\alpha)-\cot (x+\alpha)\right\}dx$\\\\\\ $=\cot(2\alpha).\int \cot (x-\alpha)dx -\cot(2\alpha).\int \cot (x+\alpha)dx$\\\\\\ $=\cot(2\alpha).\ln\mid \sin(x-\alpha) \mid - \cot(2\alpha).\ln\mid \sin(x+\alpha) \mid+C$\\\\\\ $=\cot(2\alpha).\ln \mid \frac{\sin(x-\alpha)}{\sin(x+\alpha)} \mid +C$
1708\hspace{-16}(3)::\int\frac{\left(\sec^7 x-7\right)}{\sin^2 x}dx=\int\frac{\sec^7 x}{\sin^2 x}-7.\int\frac{1}{\sin^2 x}dx\\\\\\ =\int \underbrace{\sec^7 x}_{1}.\underbrace{\csc^2 x}_{2}dx-7.\int\csc^2 xdx$\\\\\\ Apply integration By parts for First, We Get\\\\ $=\sec^7 x.(-\cot x)-7.\int\sec^6 x.\sec x.\tan x.(-\cot x)dx-7(-\cot x)+C $\\\\\\ $=-\sec^7 x.\cot x+7.\int \sec^7 xdx+7\cot x+C$\\\\ Now We Calculate value of $\int\sec^7 x dx$\\\\\\ So First We Calculate for $\int\sec^n x dx$\\\\\\ Let $I_{n}=\int\sec^n x dx=\int\sec^{n-2}x.\sec^2 xdx$\\\\\\ $I_{n}=\sec^{n-2}x.\tan x-(n-2).\int \sec^{n-3}x.\sec x.\tan x.\tan xdx$\\\\\\ $I_{n}=\sec^{n-2}x.\tan x-(n-2)\int \sec^{n-2}.(\sec^2 x-1)dx$\\\\\\ $I_{n}=\sec^{n-2}x.\tan x-(n-2).\int \sec^n xdx+(n-2).\int\sec^{n-2}xdx$\\\\\\ $I_{n}=\sec^{n-2}x.\tan x-(n-2).I_{n}+(n-2).I_{n-2}$\\\\\\ $I_{n}=\frac{\sec^{n-2}.\tan x}{n-1}+\frac{n-2}{n-1}.I_{n-2}$
\hspace{-16} $Replace $n\rightarrow (n-2)\;,$ We Get\\\\ $I_{n-2}=\frac{\sec^{n-4}.\tan x}{n-3}+\frac{n-4}{n-3}.I_{n-4}$\\\\ Again Replace $n\rightarrow (n-2)\;,$ We Get\\\\ $I_{n-4}=\frac{\sec^{n-6}.\tan x}{n-5}+\frac{n-6}{n-5}.I_{n-6}$\\\\ Now Put This value in $I_{n-2}\;,$ We Get\\\\ $I_{n-2}=\frac{\sec^{n-4}.\tan x}{n-3}+\frac{n-4}{n-3}.\left\{\frac{\sec^{n-6}.\tan x}{n-5}+\frac{n-6}{n-5}.I_{n-6}\right\}$\\\\ so $I_{n-2}=\frac{\sec^{n-4}.\tan x}{n-3}+\frac{(n-4)}{(n-3).(n-5)}.\sec^{n-6}.\tan x+\frac{(n-4).(n-6)}{(n-3).(n-5)}.I_{n-6}$\\\\\\ So put value of $I_{n-2}$ in $I_{n}\;,$ We Get\\\\ $I_{n}=\frac{\sec^{n-2}.\tan x}{n-1}+\frac{(n-2)}{(n-1).(n-3)}.\sec^{n-4}.\tan x+\frac{(n-2).(n-4)}{(n-1).(n-3).(n-5)}.\sec^{n-6}.\tan x+\\\\\frac{(n-2).(n-4).(n-6)}{(n-1).(n-3).(n-5)}.I_{n-6}$\\\\\\ Put $n=7\;,$ We Get\\\\ $I_{7}=\int \sec^7 xdx=\frac{1}{6}.\sec^5 x.\tan x+\frac{5}{24}.\sec^3 x.\tan x+\frac{5}{16}.\sec x.\tan x+\\\\\frac{5}{16}\ln\mid \sec x+\tan x \mid+C$
1708Where Ei = exponential Integral
I Think not in terms of elementry function.
1Man 111 thanks for the solutions...the first on..ok.,thts rite..!! but for the second one the options are in sin and cos and also the ans is jus a fraction!
15) ∫log(cosx) sec2x dx
just use byparts to get answer as,
tanx.log(cosx) + tanx - x
for the last question by shubham,
take log(x + ....) as t so the answer is 4pi *t22