Question

تكامل (س+جاس) ÷(١+جتاس) دس

Answer 1

sin(x)+xcos(x)+1

xcsc(x)+cot(x)+C

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Problem:

∫sin(x)+xcos(x)+1dx

Expand:

=∫(sin(x)cos(x)+1+xcos(x)+1)dx

Apply linearity:

=∫sin(x)cos(x)+1dx+∫xcos(x)+1dx

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Now solving:

∫sin(x)cos(x)+1dx

Substitute u=cos(x)+1 ⟶ dudx=−sin(x) (steps) ⟶ dx=−1sin(x)du:

=−∫1udu

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Now solving:

∫1udu

This is a standard integral:

=ln(u)

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Plug in solved integrals:

−∫1udu

=−ln(u)

Undo substitution u=cos(x)+1:

=−ln(cos(x)+1)

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Comments

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Now solving:

∫xcos(x)+1dx

Rewrite/simplify using trigonometric/hyperbolic identities:

=∫xcsc(x)csc(x)+cot(x)dx

Integrate by parts: ∫fg′=fg−∫f′g

f	=x,	g′	=csc(x)csc(x)+cot(x)
↓ steps	↓ steps
f′	=1,	g	=1csc(x)+cot(x):

=xcsc(x)+cot(x)−∫1csc(x)+cot(x)dx

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Now solving:

∫1csc(x)+cot(x)dx

Prepare for substitution, use:
csc(x)=1sin(x),
cot(x)=cos(x)sin(x)

=∫−(−1cos(x)+1)sin(x)dx

Substitute u=cos(x) ⟶ dudx=−sin(x) (steps) ⟶ dx=−1sin(x)du:

=−∫1u+1du

… or choose an alternative:
sec(x)

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Now solving:

∫1u+1du

Substitute v=u+1 ⟶ dvdu=1 (steps) ⟶ du=dv:

=∫1vdv

Use previous result:

=ln(v)

Undo substitution v=u+1:

=ln(u+1)

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Plug in solved integrals:

−∫1u+1du

=−ln(u+1)

Undo substitution u=cos(x):

=−ln(cos(x)+1)

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Plug in solved integrals:

xcsc(x)+cot(x)−∫1csc(x)+cot(x)dx

=ln(cos(x)+1)+xcsc(x)+cot(x)

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Plug in solved integrals:

∫sin(x)cos(x)+1dx+∫xcos(x)+1dx

=xcsc(x)+cot(x)

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The problem is solved:

∫sin(x)+xcos(x)+1dx

=xcsc(x)+cot(x)+C