What is $f (x) = \int x sin 4 x + 2 tan x d x$ $f(x) = int xsin4x+2tanx dx$ if $f (\frac{π}{4}) = 1$ $f(pi/4)=1$ ?

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Answer 1 · 2018-07-18T13:24:48

$F (x) = \frac{1}{16} [sin 4 x - 4 x cos 4 x] + ln ∣ ∣ {sec}^{2} x ∣ ∣ + 1 - \frac{π}{16} - ln 2$ $F(x)=1/16[sin4x-4xcos4x]+ln|sec^2x|+1-pi/16-ln2$

Here ,

$f (x) = \int x sin 4 x d x + 2 \int tan x d x = I_{1} + 2 I_{2}$ $f(x)=intxsin4xdx+2int tanxdx=I_1+2I_2$

Now,

$I_{1} = \int x sin 4 x d x$ $I_1=intxsin4xdx$

$I_{1} = x \cdot \int sin 4 x d x - \int (1 \cdot \int sin 4 x d x) d x$ $I_1=x*intsin4xdx-int(1*intsin4xdx)dx$

$\Rightarrow I_{1} = x \cdot (- \frac{cos (4 x)}{4}) - \int (- \frac{cos (4 x)}{4}) d x$ $=>I_1=x*(-cos(4x)/4)-int(-cos(4x)/4)dx$

$\Rightarrow I_{1} = - \frac{x}{4} cos 4 x + \frac{1}{4} \int cos 4 x d x$ $=>I_1=-x/4cos4x+1/4intcos4xdx$

$\Rightarrow I_{1} = - \frac{x}{4} cos 4 x + \frac{1}{4} \frac{sin (4 x)}{4} + c_{1}$ $=>I_1=-x/4cos4x+1/4sin(4x)/4+c_1$

$\Rightarrow I_{1} = \frac{1}{16} [sin 4 x - 4 x cos 4 x] + c_{1}$ $=>I_1=1/16[sin4x-4xcos4x]+c_1$

Again , $I_{2} = \int tan x d x = ln | sec x | + c_{2}$ $I_2=int tanxdx=ln|secx|+c_2$

So, $f (x) = I_{1} + 2 I_{2}$ $f(x)=I_1+2I_2$

$\Rightarrow F (x) = \frac{1}{16} [sin 4 x - 4 x cos 4 x] + 2 ln | sec x | + c_{1} + c_{2}$ $=>F(x)=1/16[sin4x-4xcos4x]+2ln|secx|+c_1+c_2$

$f (x) = \frac{1}{16} [sin 4 x - 4 x cos 4 x] + ln ∣ ∣ {sec}^{2} x ∣ ∣ + C \to C = c_{1} + c_{2}$ $f(x)=1/16[sin4x-4xcos4x]+ln|sec^2x|+C toC=c_1+c_2$

So,

$f (\frac{π}{4}) = \frac{1}{16} [sin π - π cos π] + ln ∣ ∣ {sec}^{2} (\frac{π}{4}) ∣ ∣ + C = 1$ $f(pi/4)=1/16[sinpi-picospi]+ln|sec^2(pi/4)|+C=1$

$\Rightarrow \frac{1}{16} [0 - π (- 1)] + ln ∣ ∣ {(\sqrt{2})}^{2} ∣ ∣ + C = 1$ $=>1/16[0-pi(-1)]+ln|(sqrt2)^2|+C=1$

$\Rightarrow \frac{π}{16} + ln 2 + C = 1$ $=>pi/16+ln2+C=1$

$\Rightarrow C = 1 - \frac{π}{16} - ln 2$ $=>C=1-pi/16-ln2$

Hence ,

$F (x) = \frac{1}{16} [sin 4 x - 4 x cos 4 x] + ln ∣ ∣ {sec}^{2} x ∣ ∣ + 1 - \frac{π}{16} - ln 2$ $F(x)=1/16[sin4x-4xcos4x]+ln|sec^2x|+1-pi/16-ln2$

What is f(x) = int xsin4x+2tanx dxf(x)=∫xsin4x+2tanxdxf(x) = int xsin4x+2tanx dx if f(pi/4)=1 f(π4)=1f(pi/4)=1 ?